In this proof, there is a statement where:
$$(a+b)^{p^nk}\equiv (a^{p^n}+b^{p^n})^k\mod p$$ I understand this part. But then it expands both sides binomially, and compares coefficients of $\displaystyle{a^{p^n(k-1)}b^{p^n}}$ to conclude $\displaystyle{{p^nk\choose p^n}}\equiv k\mod p$.
If $k<p$, then none of the coefficients in the binomial expansion of $(a^{p^n}+b^{p^n})^k$ are divisible by $p$. I don't understand why comparing coefficients would give us the answer here.
Thanks in advance!
The idea is that you calculate the coefficient of $a^{p^n(k-1)}b^{p^n}$ in two ways and compare the results. The equality is only a congruence modulo $p$, so the result we get is also a congruence modulo $p$.
The first way: $$(a+b)^{p^nk}=\sum_{i=0}^{p^nk}{p^nk\choose i}a^{p^nk-i}b^i.$$ To get a term $a^{p^n(k-1)}b^{p^n}$ we must select $i=p^n.$ Thus we get that $$ (a+b)^{p^nk}=\cdots+{p^nk\choose p^n}a^{p^n(k-1)}b^{p^n}+\cdots. $$
The second way: Here we calculate (after that application of Freshman's dream) $(a^{p^n}+b^{p^n})^k$. Again using the binomial theorem. This time we get $$ (a^{p^n}+b^{p^n})^k=\sum_{i=0}^k{k\choose i}a^{p^n(k-i)}b^{p^ni}. $$ To get the desired term we must select $i=1$. Thus $$ (a^{p^n}+b^{p^n})^k=\cdots+{k\choose 1}a^{p^n(k-1)}b^{p^n}+\cdots. $$ We are working in an algebraic structure (to be more precise: the ring of polynomials in two unknowns $a,b$, with coefficients in the residue class ring of integers $\mathbb{Z}_p$), where the presentation of a polynomial in terms of the basis of monomials is unique. Therefore we can conclude that in the ring $\mathbb{Z}_p$ we have $$ {p^nk\choose p^n}={k\choose 1}, $$ or, writing the same thing as a congruence, the claim $$ {p^nk\choose p^n}\equiv k\pmod p. $$
