Not exactly &quot;proportional&quot;, because you have that pesky e in the denominator.  But that&#x27;s the right idea.  Basically, to get n! you&#x27;re multiplying together n things that are sort of n-ish, so you&#x27;d expect n! ~ n^n.  (When the numbers get really big, like in statistical mechanics, I&#x27;ve seen the approximation log n! ~ n log n.)   The next step is to figure that you&#x27;re multiplying together n things that are on average n&#x2F;2, so n! ~ (n&#x2F;2)^n.  But then it really turns out that you should have been using a geometric average (since you&#x27;re multiplying), not an arithmetic one, so n! must be smaller yet.  (I don&#x27;t know a way to get (n&#x2F;e)^n without doing an integral, though.)