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 1:

 $$(a+1)^p=\sum_{i=0}^{p}a^i\binom{p}{i}$$

Hint 2:

 What is $\binom{p}{i}~\text{mod}~p$?

