Binomial Coefficient involving Power of Prime/Proof 1

Proof
From Prime Power of Sum Modulo Prime we have:
 * $(1): \quad \paren {a + b}^{p^n} \equiv \paren {a^{p^n} + b^{p^n} } \pmod p$

We can write this:
 * $\paren {a + b}^{p^n k} = \paren {\paren {a + b}^{p^n} }^k$

By $(1)$ and Congruence of Powers, we therefore have:
 * $\paren {a + b}^{p^n k} \equiv \paren {a^{p^n} + b^{p^n} }^k \pmod p$

The coefficient $\dbinom {p^n k} {p^n}$ is the binomial coefficient of $b^{p^n}$ in $\paren {a + b}^{p^n k} = \paren {\paren {a + b}^{p^n} }^k$.

Expanding $\paren {a^{p^n} + b^{p^n} }^k$ using the Binomial Theorem, we find that the coefficient of $b^{p^n}$, the second term, is $\dbinom k 1 = k$.

So:
 * $\dbinom {p^n k} {p^n} \equiv k \pmod p$