Talk:Binomial Coefficient involving Power of Prime/Proof 1

I think the conclusion of this proof can be slightly expanded

Specifically this part:

"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, $\dbinom {p^n k} {p^n} \equiv k \pmod p$".

I'm a little confused about how the result follows from this part. Substituting $1$ for $a$, I can see that $\ds\sum_{i=0}^{p^nk}\dbinom {p^n k} {i} b^i \equiv \ds\sum_{i=0}^{k} \dbinom {k} {i} b^{ip^n}\pmod{p}$. So $b^{p^n}$ will have coeffiecients of $\dbinom {p^n k} {p^n}$ and $\dbinom k 1$ on the left and right side of the congruence, respectively. The expansion on the left has $p^nk+1$ terms and the expansion on the right has $k+1$ terms, so you can't really cancel anything out. Is there some property of congruences or modular arithmetic being used here?