Prove that the limit definition of the exponential function implies its infinite series definition. Here's the problem: Let $x$ be any real number. Show that
$$
\lim_{m \to \infty} \left( 1 + \frac{x}{m} \right)^m = \sum_{n=0}^ \infty \frac{x^n}{n!} 
$$
I'm sure there are many ways of pulling this off, but there are 3 very important hints to complete the exercise in the desired manner:


*

*Expand the left side as a finite sum using the Binomial Theorem. Call the summation variable $n$.

*Now add into the finite sum extra terms which are $0$ for $n>m$, in order to make it look like an infinite series.

*What happens to the limit on $m$ outside the series?


So far I was able to use Hint 1 to expand the left side:
$$
\lim_{m \to \infty} \left( 1 + \frac{x}{m} \right)^m = \lim_{m \to \infty} \sum_{n=0}^m \binom {m}{n} \left( \frac{x}{m} \right)^n
$$
No matter what I do with the binomial coefficients and factorials, I can't figure out what extra terms to add per Hint 2. Any suggestions?

Recall that
$$\binom{m}{n}=\frac{m!}{n! (m-n)!}$$
By Stirling's formula you get for large $m$
$$m!\approx \sqrt{2\pi m}\left(\frac{m}{e}\right)^{m}\\
(m-n)!\approx \sqrt{2\pi(m-n)}\left(\frac{m-n}{e}\right)^{m-n} $$
Then note
$$ \left(\frac{m-n}{e}\right)^{m-n}=e^{(m-n)\ln(m-n)-m+n}=e^{(m-n)\ln(m)+(m-n)\ln(1-\frac{n}{m})-m+n} $$
By Taylor series
$$f(x+\epsilon)=f(x)+f'(x)\epsilon+O(\epsilon^2)~~~\text{get}~~~\ln(1-\frac{n}{m})=\underbrace{\ln(1)}_{=0}-\frac{n}{m}+O(m^{-2})$$
and we proceed to
$$ \left(\frac{m-n}{e}\right)^{m-n}\approx e^{(m-n)\ln(m)-(m-n)\frac{n}{m}-m+n}\approx e^{(m-n)\ln(m)-m}= \frac{m^{m-n}}{e^m}$$
where we neglected terms that certainly vanish in the limit $m\to\infty$.
Therefore, we conclude that
$$\frac{m!}{(m-n)!}= m^n+\text{subleading}$$
And all together
$$\sum_{n=0}^\infty \binom{m}{n} \left(\frac{x}{m}\right)^n= \sum_{n=0}^\infty \frac{m^n}{n!} \left(\frac{x}{m}\right)^n+\text{subleading}\to e^x$$
