Can one define the $p$-adic exponential function $\exp_p (x)$ as the limit of $(1 + x/n)^n$ when $n\to\infty$? I limit myself in following considerations to the field $\mathbb Q_p$. As everybody interested in ultrametric calculus knows, the $p$-adic exponential function defined by series $$\exp_p(x) = \sum_{k=0}^\infty \frac{x^k}{k!}$$ converges only on a small ball $\mathcal B$ around $0$ equal to $p\mathbb Z_p$ (for $p \ge 3$) or $4 \mathbb Z_2$ (otherwise). Inspired by my previous post ,,About $\lim \left(1+\frac {x}{n}\right)^n$'' I have decided to check whether this alternative representation is also true here, in $p$-adic context.
As a side note, an analytical function $f$ from convex neighbourhood of zero satisfying $f = f'$ or $f(x+y) = f(x) f(y)$ must be identically zero (this is exercise 25.J from Schikhof's book).
Let's fix $\varepsilon > 0$ and $x \in \mathcal B$. We have $$\left|\exp_p x - \Bigl(1 + \frac xn\Bigr)^n\right| = \left(\sum_{k=n}^\infty \frac{x^k}{k!}\right) + \sum_{k=2}^n \left(\frac{x^k}{k!} - {n \choose k} \frac{x^k}{n^k}\right).$$
because $(1 +x/n)^n = 1 + x + \ldots$.
For sufficiently large values of $n$, the left sum can be made smaller than $\varepsilon / 2$. Now I have tried to rewrite the summand as $$\frac{x^k}{k!} \cdot \left(1-\frac{n!}{(n-k)! \cdot n^k}\right)$$
and assumed that $n$ is of the form $p^m + p^{m-1} + \ldots + 1$, which implies $v_p(n!) =n - p^m$ by the Legendre formula and $|n|_p^k = 1$ by direct check. Now $v_p((n-k)!)$ is certainly not larger than $p^m / (p-1)$ and therefore $$v_p(n!) - v_p(k!) \ge \frac{p-2}{p-1} \cdot p^{m-1} + p^{m-2} + \ldots + 1 \to \infty.$$
It seems that $\exp_p (x) = \lim_{n \to \infty} (1 + x/n)^n$. Am I right?

No, the limit exists only for $x=0$
If $(a_n)$ is a sequence in $\Bbb Q_p$ and $(a_n)$ converges to anything $p$-adically, then the set $\{v_p(a_n)\}$ has a lower bound.
However, here if $x$ is nonzero and you pick an $n$ such that $v_p(n) > v_p(x)$, then $v_p(1+x/n)< 0$, and since $v_p((1+x/n)^n)$ is $n$ times that, this shows that $v_p((1+x/n)^n)$ can take arbitrarily big negative values. This prevents the sequence form converging to anything.
