Not exactly "proportional", because you have that pesky e in the denominator. But that's the right idea. Basically, to get n! you're multiplying together n th

Not exactly "proportional", because you have that pesky e in the denominator. But that's the right idea. Basically, to get n! you're multiplying together n things that are sort of n-ish, so you'd expect n! ~ n^n. (When the numbers get really big, like in statistical mechanics, I've seen the approximation log n! ~ n log n.) The next step is to figure that you're multiplying together n things that are on average n/2, so n! ~ (n/2)^n. But then it really turns out that you should have been using a geometric average (since you're multiplying), not an arithmetic one, so n! must be smaller yet. (I don't know a way to get (n/e)^n without doing an integral, though.)