How to reduce the congruence $(p-1)^{p-1}\equiv 2^{p-1}\pmod{p^2}$ into $p(p-1)+2^{p-1}-1\equiv 0\pmod{p^2}$?

Question

In addition, how is $(p-a)^{p-1}-a^{p-1}\equiv -(p-1)pa^{p-2}\pmod{p^2}$ derived by binomial theorem?

Answer 1

Note that, by the binomial theorem $$(p-1)^{p-1}=(1-p)^{p-1}=1-(p-1)p+\binom{p-1}2p^2-\cdots$$ (at least if $p$ is odd).

Answer 2

Note that $(p-1)^{p-1}$ = $p^{p-1} - (p-1)p^{p-2} + ... - (p-1)p^1 + 1$, and only the last two cannot be divided by $p^2$ , so you are left with $-(p-1)p^1 + 1 \equiv 2^{p-1}$, so $0 \equiv 2^{p-1} +p(p-1) - 1$. As for the second thing, you have $(p-a)^{p-1}$ = $p^{p-1} - (p-1)p^{p-2}a^1 + ... -(p-1)p^1a^{p-2} + a^p$, and you can proceed as before.