Consider the fact that .
Then note that because p is prime neither k! nor (p-k)! is a factor of p.
So p is a factor of .
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Can you see how to complete the argument?
Show that if p is a prime, then (a + b)^p ≡ a^p + b^p (mod p)
(Hint: prove and use that p choose k is a multiple of p for every k ∈ {1, . . . , p − 1}). Then expand (a + b)^p via the binomial theorem).
Using the previous formula, prove Fermat’s little Theorem (in its equivalent form, a^p ≡ a (mod p)) by induction on a.
Any help appreciated!