= Solution
The <Mahler theorem> states that a function $f:\mathbb Z_p\to\mathbb Q_p$ is continuous if and only if it has a unique expansion
$$
f(x)=\sum_{n\ge0}c_n\binom xn,\qquad c_n\in\mathbb Q_p,\quad c_n\longrightarrow0,
$$
and the expansion converges uniformly. Moreover $\|f\|_\infty=\sup_n|c_n|_p$; in particular $f$ takes values in $\mathbb Z_p$ exactly when all $c_n$ belong to $\mathbb Z_p$. Here $\binom{x}{0}=1$ and $\binom{x}{n}=x(x-1)\cdots(x-n+1)/n!$. These <binomial polynomials> are continuous and integer-valued on $\mathbb Z_p$, since they are integer-valued on the dense nonnegative integers. The coefficient condition $c_n\to0$ therefore makes the series uniformly convergent.
Evaluate at a nonnegative integer $m$. Terms with $n>m$ vanish, giving the finite triangular relation $f(m)=\sum_{n=0}^m\binom mn c_n$. <Binomial inversion> gives the <Mahler coefficients>
$$
\boxed{c_n=\sum_{j=0}^n(-1)^{n-j}\binom njf(j)=(\Delta^nf)(0),\qquad\Delta f(x)=f(x+1)-f(x).}
$$
To obtain the generating function, multiply the last finite identity by $T^n/n!$ and compare coefficients. Setting $n=j+k$ yields
$$
\sum_{n\ge0}\frac{c_n}{n!}T^n=\sum_{j,k\ge0}\frac{(-1)^k f(j)}{j!k!}T^{j+k}=\boxed{e^{-T}\sum_{j\ge0}\frac{f(j)}{j!}T^j.}
$$
The <Mahler coefficient exponential generating function> is an identity in $\mathbb Q_p[[T]]$: each coefficient is a finite sum. It does not assert convergence for every $p$-adic substitution for $T$.
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