Notice that if $p^{2}|a^{p-1}-1$,we have $(p-a)^{p-1}\equiv a^{p-1}-1-(p-1)pa^{p-2}\equiv a^{p-1}-1+pa^{p-2}\equiv pa^{p-2}\pmod {p^{2}}$,so $p^{2}\nmid (p-a)^{p-1}$.
Then we assume that $\forall 1\le a\le p-2$,$p^{2}|a^{p-1}-1$ or $p^{2}|(a+1)^{p-1}-1$.Because of $p^{2}|1^{p-1}-1$,$p^{2}\nmid (p-1)^{p-1}-1\Rightarrow p^{2}|(p-2)^{p-1}-1$,thus $p^{2}|2^{p-1}-1+p2^{p-2}$ $\Rightarrow p^{2}|(2^{p-1}+1)(2^{p-1}-1+p2^{p-2})=4^{p-1}-1+p2^{p-2}(2^{p-1}+1)\cdots (1)$.
Meanwhile,$p^{2}\nmid 2^{p-1}-1\Rightarrow p^{2}|3^{p-1}-1\Rightarrow p^{2}\nmid (p-3)^{p-1}-1\Rightarrow p^{2}|(p-4)^{p-1}-1$,also $p^{2}|4^{p-1}-1+p4^{p-2}\cdots (2)$.
Compare $(1)$ with $(2)$,we have $p|4^{p-2}+2^{p-2}$,but $4(4^{p-2}+2^{p-2})=4^{p-1}+2\times 2^{p-1}\equiv 3\pmod p$,contradiction!We have done the problem.