Can $e^x$ be expressed as a linear combination of $(1 + \frac x n)^n$? Can $e^x$ be expressed as a linear combination of $(1 + \frac x n)^n$? In other words, does there exist an infinite sequence $(a_k)_{k \in \mathbb N_0}$ such that $$e^x = a_0 + \sum_{1 \leq k < \infty} a_k \left(1 + \frac x k\right)^k$$
for all $x \in \mathbb R$? Call the series on the right $s(x)$.
I can answer the question in the negative when the series is absolutely convergent. In the conditionally convergent case, I'm not so sure. My thoughts were to use the fact that:
$$e^{x - \frac{x^2}{2k}} \leq (1+ \frac x k)^k \leq e^{x}$$
and use the lower bound when $a_k$ is negative, and the upper bound when $a_k$ is positive. This gets stuck because it's not always the case that if some $b_k$ is a decaying sequence then $\sum_{k} \frac{b_k}{k}$ is convergent.
The strengthened inequality $$e^{x - \frac{x^2}{2k}} \leq (1+ \frac x k)^k \leq e^{x - \frac{x^2}{2k} + \frac{x^3}{3k^2}}$$ looks like it might make more progress...

[EDIT 2019/08/14 14:00 GMT]
This is the solution in the absolutely convergent case, given by lemmas 1 and 2:

Definition: Let $s(x) = a_0 + \sum_{1 \leq k < \infty} a_k \left(1 + \frac x k\right)^k$.

Lemmas and proofs follow:

Lemma 1: If $s(x)$ converges absolutely for some $x\geq 0$, then $s(x)$ converges absolutely for all $x \geq 0$.

Proof
Pick an $x_0 \geq 0$ for which $s(x_0)$ converges absolutely.
By the condition stated in the lemma, the series $\sum_{1 \leq k < \infty} |a_k| \left|1 + \frac {x_0} k\right|^k$ must converge. We also observe that $|a_k| \leq |a_k| \left|1 + \frac {x_0} k\right|^k$ is true for all $k$. So by the Direct Comparison Test, the series $\sum_{0 \leq k < \infty} |a_k|$ must also converge. In other words, $s(0)$ is absolutely convergent.
Consider now any $x \geq 0$. The series $\sum_{0 \leq k < \infty} |a_k| e^{x}$ converges because it is equal to $e^{x} \sum_{0 \leq k < \infty} |a_k|$, which we proved to be convergent in the previous paragraph. We observe that $|a_k| \left|1 + \frac {x} k\right|^k \leq |a_k| e^{x}$ is true for all $k$. So by the Direct Comparison Test, the series $|a_0| + \sum_{1 \leq k < \infty} |a_k| \left|1 + \frac {x} k\right|^k$ must also converge. So by the definition of absolute convergence, we have that $a_0 + \sum_{1 \leq k < \infty} a_k \left(1 + \frac x k\right)^k=s(x)$ converges absolutely, where $x \geq 0$ was arbitrary.
$\blacksquare$

Lemma 2: If $s(x)$ converges absolutely when $x \geq 0$, then for large enough $x$ we have that $e^x > s(x)$.

Proof
Let $z_n(x) = |a_0| + \sum_{1 \leq k < n} |a_k| \left(1 + \frac x k\right)^k$.
Pick some $\epsilon < \frac 1 2$.
Observe that there must be a large enough $n$ such that $z_\infty(0) - z_n(0) \leq \epsilon$.
Using the triangle inequality, we have that:
$$\begin{aligned}
|s(x)| &\leq z_\infty(x)\\
&\leq z_n(x) + (z_\infty(x) - z_n(x))\\
\end{aligned}$$
Since $z_n(x)$ is a polynomial, there is a large enough $X$ such that all $x \geq X$ it's true $z_n(x) < \epsilon \cdot e^x$. So we have that
$$\begin{aligned}
|s(x)| &<\epsilon\cdot e^x + (z_\infty(x) - z_n(x))\\
&\leq \epsilon\cdot e^x + (z_\infty(0) - z_n(0)) e^x\\
&\leq \epsilon\cdot e^x + \epsilon\cdot e^x\\
& = 2\epsilon \cdot e^x\\
&< e^x.
\end{aligned}$$
The claim above that $z_\infty(x) - z_n(x) \leq (z_\infty(0) - z_n(0)) e^x$ follows from 
$$\begin{aligned}
&|a_k| \left(1 + \frac x k\right)^k \leq |a_k| e^x\\
\implies &\sum_{k \geq {n+1}}\left(1 + \frac x k\right)^k \leq \sum_{k \geq {n+1}}|a_k| e^x\\
\implies & z_\infty(x) - z_n(x) \leq (z_\infty(0) - z_n(0)) e^x
\end{aligned}$$
We are done.
$\blacksquare$

Let $R_n(z)=\sum_{1 \leq k \le n} a_k \left(1 + \frac z k\right)^k$. Assuming the hypothesis we will show:
1: $R_n(z) \to e^z-a_0$ uniformly in the disc $|z| \le \frac{1}{2}$
2: $\sum_{1}^{\infty}{\frac{a_k}{k^q}}=0, q \ge 1$ arbitrary integer
3: $a_k=0, k \ge 1$
We use that if $|z| \le \frac{1}{2}, k \ge 1$, $|(1+\frac{z}{k})^k-(1+\frac{z}{k+1})^{k+1}| \le B|\frac{1}{k^2}|, B>0$ constant, which follows from $k\log(1+\frac{z}{k})=z-\frac{z^2}{2k}+O(\frac{z^3}{k^2}), |z| \le \frac{1}{2}, k \ge 1$, so $(1+\frac{z}{k})^k=e^{z-\frac{z^2}{2k}+O(\frac{z^3}{k^2})}$ and then subtracting the relations for $k, k+1$ since the $O$ terms are at most $\frac{1}{8k^2}$ in absolute value and $|\frac{z^2}{2k}-\frac{z^2}{2k+2}| \le \frac{1}{8k^2}, |z| \le \frac{1}{2}$
We also note that $|(1+\frac{z}{k})^k| \le (1+\frac{|z|}{k})^k \le e^{|z|} \le e^{\frac{1}{2}} <e$ since the binomial coefficients are positive and the triangle inequality works
For simplicity let $b_k(z)=(1+ \frac z k)^k$, so if $|z| \le \frac{1}{2}$
$|b_k(z)-b_{k+1}(z)| \le B|\frac{1}{k^2}|$
$|b_k(z)| \le e$
Then since $\Sigma{a_k} \to 1$ by hypothesis for $x=0$, it folows that $|\sum_{N}^{M}a_k| \le A$ for all $N \le M$ and some constant $A>0$, while $|\sum_{N}^{M}a_k| \to 0, N,M \to \infty$ so if we pick arbitrary $\epsilon >0, |\sum_{N}^{M}a_k| \le \epsilon, M>N >N(\epsilon)$ and then we sum by parts:
$|R_M(z)-R_N(z)|=|\sum_{N+1}^{M}a_kb_k(z)|=|(A_{N+1}(b_{N+1}-b_{N+2})(z))+(A_{N+2}(b_{N+2}-b_{N+3})(z))+....(A_{M-1}(b_{M-1}-b_{M})(z))+(A_{M}(b_{M}(z))|$, 
where $A_p=\sum_{N+1}^{p}a_k, p \ge N+1$
So $|R_M(z)-R_N(z)| \le A\sum_{N+1}^{M-1}{B|\frac{1}{k^2}}|+e|A_M| \le AB\frac{1}{N}+e\epsilon, M>N > N(\epsilon)$ which shows that $R_N(z)$ is uniformly Cauchy in $|z| \le \frac{1}{2}$. But this means $R_n(z)$ converges uniformly to an analytic function $f(z)$ on the disc of radius $r=\frac{1}{2}$ and since we know by hypothesis that $f(x)=e^x-a_0$ on the $[-\frac{1}{2},\frac{1}{2}]$ segment it follows by the identity principle that $f(z)=e^z-a_0$ and this is 1 above
Now, we can differentiate term by term and get $R_n(z)' \to e^z$ uniformly on the  disc of radius $\frac{1}{2}$ and then plugging $z=0$ we get $\sum_{k \ge 1}a_k=1$ hence $a_0=0$, hence $R_n(z) \to e^z$ uniformly on the above disc. 
Subtracting we get $\sum_{k = 1}^n\frac{a_kz}{k}(1+\frac{z}{k})^{k-1} \to 0$ uniformly which clearly implies $\sum_{k = 1}^n\frac{a_k}{k}(1+\frac{z}{k})^{k-1} \to 0$, hence $\sum_{k \ge 1}\frac{a_k}{k}=0$
($zf_n(z) \to 0$ uniformly on the disc of radius $r$, means that for any $\epsilon >0$ there is $N(\epsilon), |zf_n(z)| \le \epsilon, |z| \le r, n \ge N(\epsilon)$ 
Schwarz lemma implies $|zf_n(z)| \le \frac{|z|}{r}\epsilon, |z| \le r, n \ge N(\epsilon)$ or $|f_n(z)| \le \frac{1}{r}\epsilon, |z| \le r, n \ge N(\epsilon)$)
But now (with all convergences below being uniform) we can integrate on the straight line from $0$ to $z$, $\sum_{k = 1}^{n}\frac{a_k}{k}(1+\frac{w}{k})^{k-1} \to 0$ and get $\sum_{k = 1}^{n}\frac{a_k}{k}(1+\frac{z}{k})^{k} \to 0$. 
Subtracting gives $\sum_{k = 1}^n\frac{a_kz}{k^2}(1+\frac{z}{k})^{k-1} \to 0$, hence $\sum_{k = 1}^n\frac{a_k}{k^2}(1+\frac{z}{k})^{k-1} \to 0$, hence $\sum_{k \ge 1}\frac{a_k}{k^2}=0$. A clear induction (integrate, subtract, divide the $z$) gives 2 above
3 is a trivial consequence of 2 since wlog we can assume $\sum|a_k| < \infty$ in 2 by going to $b_k=\frac{a_k}{k^2}$ which is absolutely convergent since $a_k$ is bounded and which clearly satisfies 2; then if $p \ge 1$ is the first index for which $a_p \ne 0, |a_p|=a>0$ and with $A=\max|a_k| \ge a>0$ we easily find a large $q$ s.t. $A\frac{p^q}{(p+m)^q} \le .0001\frac{a}{(p+m)^2}, m>1$ as $(1+\frac{m}{p})^{q-2} \ge (1+\frac{1}{p})^{q-2} \to \infty$ with $q$ for fixed $p$, leading to a term with absolute value $a$ plus sum that is at most $.0001a\frac{\pi^2}{6}$ in absolute value being zero and that is impossible. 
So all $a_k$ must be zero if 2 is satisfied and we are finally done!
