TITLE: Logic behind the multiplicative form of "Gauss trick"?
QUESTION [5 upvotes]: I was going through $ \lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite
I don't follow why/how this is true $(n!)^2 = (1 \cdot n) (2 \cdot (n-1)) (3 \cdot (n-2)) \cdots ((n-2) \cdot 3) ((n-1) \cdot 2) (n \cdot 1)\ge n^n$
I read that At least half the terms in the product for n! will be at least n/2 .. so ?
Please advise.

REPLY [5 votes]: $$(n!)^2 = \overbrace{(1 \cdot n)}^{\ge n} \overbrace{(2 \cdot (n-1))}^{\ge 2\cdot n/2} \overbrace{(3 \cdot (n-2))}^{\ge 2\cdot n/2} \cdots \overbrace{((n-2) \cdot 3)}^{\ge 2\cdot n/2} \overbrace{((n-1) \cdot 2)}^{\ge 2\cdot n/2} \overbrace{(n \cdot 1)}^{\ge n}\ge n^n$$