Just A Tad Ironic

A rather trivial fact (from an exercise in Lang's Undergraduate Algebra), but interesting, all the same.

Typically, when faced with expanding the expression (x + y)^p, where p is a positive integer, many a not-thinking student will write:
(x + y)^p = x^p + y^p
(to the despair of markers, I must say.)
The ironic thing is, it's true! Sometimes. Namely, when x, y, are integers, p prime. And taking the expression mod p.

Claim:
(x + y)^p (mod p) = (x^p + y^p) (mod p)
Proof:
First, a lemma.

Lemma:
For all primes p, integers 0 < n < p, p | _pC_n (p choose n).
Proof:
By definition,
_pC_n = p!/[n!(p-n)!] = [p*(p-1)*...*(p-n+1)]/n!
Since p is prime and 0 < n < p, n! does not have p as a factor, i.e. gcd(n!, p) = 1. Since _pC_n is an integer, we must have
n! | [(p-1)*...*(p-n+1)]
(say n!a = [(p-1)*...*(p-n+1)], where a is a positive integer), which means that
_pC_n = pa
which is divisible by p.

Now, by the binomial theorem and our lemma,

(x + y)^p (mod p)
= _pC₀x^py⁰ + _pC₁x^p-1y¹ + ... + _pC_p-1x¹y^p-1 + _pC_px⁰y^p (mod p)
= x^p + 0 + ... + 0 + y^p (mod p)
= x^p + y^p (mod p)

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