Show that for any integer a and prime p, $(a+1)^p \equiv a^p+ 1 \pmod{p}$. I believe that this may require the use of Fermat's Little Theorem. I rewrote it as $(a+1)^p - a^p \equiv 1 \pmod{p}$ because the right-hand side looks similar to Fermat's Little Theorem, but I was unable to figure out how I can get the left-hand side to become $a^{(p-1)}$.  

Hint $ $ Use little Fermat to show $\,\color{#c00}n^p \equiv \color{#c00}n\pmod{\!p}\, $ for all $\,n\,$ (separate into cases $\,n\equiv 0\,$ or not), then  use that to show $\,(a+1)^p$ and $\,a^p+1\,$ are both $\,\equiv a+1\pmod{\!p}\ $ [by specializing $\,\color{#c00}n \equiv a+1\,$ and $\,\color{#c00}n\equiv a,\,$ above, while using the Congruence Power Rule].
