Binomial Coefficient involving Power of Prime/Proof 1

Lemma
Let $p$ be a prime number, and let $k \in \Z$.

Then:
 * $\displaystyle \binom {p^n k} {p^n} \equiv k \pmod p$

where $\displaystyle \binom {p^n k} {p^n}$ is a binomial coefficient.

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

We can write this:
 * $\left({a + b}\right)^{p^n k} = \left({\left({a + b}\right)^{p^n}}\right)^k$

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

The coefficient $\displaystyle \binom {p^n k} {p^n}$ is the binomial coefficient of $b^{p^n}$ in $\left({a + b}\right)^{p^n k} = \left({\left({a + b}\right)^{p^n}}\right)^k$.

Expanding $\left({a^{p^n} + b^{p^n}}\right)^k$ using the Binomial Theorem, we find that the coefficient of $b^{p^n}$, the second term, is $\displaystyle \binom {k} {1} = k$.

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