An application of this theorem to $p$-adic analysis is the $p$-integrality of the coefficients of the Artin-Hasse exponential $$ {\rm AH}_p(X) = e^{X + X^p/p + X^{p^2}/p^2 + \cdots}. $$ That means when this series is expanded as $\sum_{k \geq 0} a_kX^k$, the coefficients $a_k$ don't have their denominator divisible by $p$. This is obvious for $k = 0$ since $a_0 = 1$. We have $a_k = 1/k!$ for $1 \leq k \leq p-1$, so $a_1,\dots,a_{p-1}$ are all $p$-integral. And $a_p = ((p-1)!+1)/p!$, which is $p$-integral by Wilson's theorem. To handle the general case we'll use the Frobenius theorem.

For $k \geq 1$, a non-obvious combinatorial/probabilistic formula for the coefficients is $$ a_k = \frac{\#\{g \in S_k : g \text{ has } p\text{-power order}\}}{k!}. $$ Using the Frobenius theorem with $n$ being the largest power of $p$ dividing $|S_k|$, the numerator of $a_k$ is divisible by the largest power of $p$ in the denominator. Hence each $a_k$ is $p$-integral.

There are simpler proofs of the $p$-integrality of the coefficients of this series. See the Wikipedia page on the Artin-Hasse exponential.

An application of this theorem to $p$-adic analysis is the $p$-integrality of the coefficients of the Artin-Hasse exponential $$ {\rm AH}_p(X) = e^{X + X^p/p + X^{p^2}/p^2 + \cdots}. $$ That means when this series is expanded as $\sum_{k \geq 0} a_kX^k$, the coefficients $a_k$ don't have their denominator divisible by $p$. This is obvious for $k = 0$ since $a_0 = 1$. We have $a_k = 1/k!$ for $1 \leq k \leq p-1$, so $a_1,\dots,a_{p-1}$ are all $p$-integral. And $a_p = ((p-1)!+1)/p!,$ which is $p$-integral by Wilson's theorem. To handle the general case we'll use the Frobenius theorem.

For $k \geq 1$, a non-obvious combinatorial/probabilistic formula for the coefficients is $$ a_k = \frac{\#\{g \in S_k : g \text{ has } p\text{-power order}\}}{k!}. $$ Using the Frobenius theorem with $n$ being the largest power of $p$ dividing $|S_k|$, the numerator of $a_k$ is divisible by the largest power of $p$ in the denominator. Hence each $a_k$ is $p$-integral.

There are simpler proofs of the $p$-integrality of the coefficients of this series. See the Wikipedia page on the Artin-Hasse exponential.