Remember the Wallis product?

π/2 = 2/1 · 2/3 · 4/3 · 4/5 · 6/5 · 6/7 · ...

We can rewrite that in a really neat way. Take the partial product:

2/1 · 2/3 · 4/3 · 4/5 · 6/5 · 6/7 · ... · (2N)/(2N−1) · (2N)/(2N+1)

Divide numerators and denominators by two:

1/0.5 · 1/1.5 · 2/1.5 · 2/3.5 · 3/3.5 · 3/4.5 · ... · N/(N−½) · N/(N+½)

Divide them into two groups, alternatingly:

[1/0.5 · 2/1.5 · 3/2.5 · ... N/(N−½)] ·
[1/1.5 · 2/2.5 · 3/3.5 · ... N/(N+½)]

The numerator is the product from 1 to N twice, so it's (N!)^(2).

@LeakyNun Let me write down a very precise story for the separability stuff without the scheme-theoretic stuff anyway. Suppose $F$ is a field of characteristic $p$ such that $\alpha \in F \setminus F^p$. Then $x^p - \alpha$ is irreducible in $F[x]$ because in the algebraic closure of $F$, $x^p - \alpha = (x - \alpha^{1/p})^p$, so by unique factorization in $\overline{F}[x]$, any factor must be of the form $(x - \alpha^{1/p})^n$ where $n < p$ but this cannot be in $F[x]$ because $\alpha^{n/p} \notin F$ since if not, we take $m$ such that $mn \equiv 1 \pmod{p}$, and take the $m$-th power to o…