I'm trying to learn number theory on my own, and here's a proof I'm not quite sure I got right. It feels too simple(?), I'm thinking maybe I'm missing something. So the question is:
Prove that if $n$ is a square, then each exponent in its prime-power decomposition is even.
My proof:
Let $n=m^2$, with $m$ having prime factors $p_i$ with exponents $e_i$ so that
$m= p^{e_1}_1 p^{e_2}_2 \ldots p^{e_n}_n.$
When squared, this gives
$m^2 = (p^{e_1}_1 p^{e_2}_2 \ldots p^{e_n}_n)^2 = p^{2e_1}_1 p^{2e_2}_2 \ldots p^{2e_n}_n,$
where all the exponents are even.