I am trying to find $\lim \limits_{n \to \infty}{(n!)^n \over n^{n^2}}$. I rewrite the expression as $$\left({n! \over n^n}\right)^n = \left({1*2*\dots*n \over n*n*\dots*n}\right)^n = \left({1 \over n}\right)^n\left({2 \over n}\right)^n\dots\left({n \over n}\right)^n \le 1$$ It seems obvious that taking the limit will result in $n$ limits, all but the last one equal $0$. At the same time, this feels somewhat informal as we haven't rigorously concluded about the influence of powers on the terms. What would be a more formal way to approach this problem?
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The number of factors on the RHS depends on $n$. So you cannot take the limit on the right side as product of the limits because you would have a variable number of factors.
Correct approach: $0<\left(\frac{n!}{n^n}\right)^n\le \frac{1}{n^n}$ (the rest of the factors are $\le 1$) and use squeeze theorem.
