I want to show that the sequence $e_n = \sum_{i = 0} \frac{x^i}{i!}$ is Cauchy with respect to the p-adic norm when $p > 2$ and $|x|_p < 1$. So far, I have the estimate $|e_n - e_m|_p = |\sum_{i = m+1}^n \frac{x^i}{i!}|_p \leq \text{max}_{m + 1 \leq i \leq n}(|\frac{x^i}{i!}|_p) = |\frac{x^j}{j!}|_p$ for some $j$ in the suitable range. Just going by the definitions, we have that $|\frac{x^j}{j!}|_p < \frac{1}{|j!|_p} = p^{\text{ord}_p(j!)}$. I'm not sure how I can show how $p^{\text{ord}_p(j!)}$ becomes arbitrarily small when $m$ and $n$ (and subsequently $j$) are bounded below by a sufficiently high $M$. Likewise, I don't see how to utilize the fact that $p > 2$.
After this, I need to show that the sequence is Cauchy if $p = 2$ and $|x|_p < \frac{1}{2}$.
In fact, the radius of convergence of the exponenntial series is $\displaystyle R_p=p^{-1/(p-1)}<1$ in the $p$-adic setting ($x\in \mathbb{C}_p$, the completion of an algebraic closure of $\mathbb{Q}_p$). This can be shown from Legendre's formula (see https://en.wikipedia.org/wiki/Legendre's_formula).
So if $R_p<|x|_p<1$, your series does not converge. Now if you suppose that $x\in \mathbb{Q}_p$, the fact that $|x|_p<1$ imply $|x|_p\leq p^{-1}$. So you have $\displaystyle |\frac{x^j}{j!}|_p\leq p^{-j}p^{\frac{j-S_p(j)}{p-1}}$, (with $S_p(j)$ the sum of the digits of $j$ in its $p$-adic expansion) and with $p>2$, you can show your assertion.