Binomial Coefficient involving Power of Prime

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

Then $$\binom {p^n k} {p^n} \equiv k \left({\bmod\, p}\right)$$, where $$\binom {p^n k} {p^n}$$ is a binomial coefficient.

Proof
We have, from Binomial Coefficient of Prime: $$\forall k \in \mathbb{Z}: 0 < k < p: \binom p k \equiv 0 \left({\bmod\, p}\right)$$.

Let $$P$$ be the product of any number of the above binomial coefficients.

Then from Congruence of Product, $$P \equiv 0 \left({\bmod\, p}\right)$$.

So from Prime Power of Sum Modulo Prime we have: $$\left({a + b}\right)^{p^n} \equiv \left({a^{p^n} + b^{p^n}}\right) \left({\bmod\, p}\right)$$

as all binomial coefficients apart from the first and last are like $$P$$ above.

Thus $$\left({a + b}\right)^{p^n k} \equiv \left({a^{p^n} + b^{p^n}}\right)^k \left({\bmod\, p}\right)$$

The coefficient $$\binom {p^n k} {p^n}$$ is the coefficient of $$x^{p^n}$$ in the left hand side of the above, and congruent modulo $$p$$ to the coefficient of $$x^{p^n}$$ on the right hand side.

Thus it's congruent to $$k$$ modulo $$p$$.

Comment
This lemma is used in the proof of Sylow's Theorems.