Convergence (or not) of $\sum_{n=1}^{\infty}\frac{1!2!\cdots n!}{n^n}$

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The title says for itself: Does the series $\sum_{n=1}^{\infty}\frac{1!2!\cdots n!}{n^n}$ converges? So, as soon as I've saw the factorial, could not help but use the ratio test: \begin{align*} \lim \left|\frac{\frac{1!\cdots (n+1)!}{(n+1)^{(n+1)}}}{\frac{1!\cdots n!}{n^n}}\right| = \lim \left|\frac{n!n^n}{(n+1)^n}\right| = \lim n! \lim \left(\frac{n}{n+1}\right)^n = \text{diverges}. \end{align*} So, the series itself diverges. Is it right? I mean, looks OK, but I'm not used to deal with products inside series.

Any help would be very appreciated! Thanks in advance!

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Some comments are confusing the OP.

$$\frac{\frac{1!2!\cdots(n+1)!}{(n+1)^{n+1}}}{\frac{1!2!\cdots n!}{n^n}} = \frac{1!2!\cdots n!(n+1)! n^n}{1!2!\cdots n!(n+1)^{n+1}} = \frac{(n+1)! n^n}{(n+1)^{n+1}} = \frac{n!n^n}{(n+1)^n} = n!\left(\frac{n}{n+1}\right)^n$$

as originally the OP calculated, not as the comments suggest.

So yes, your solution is correct.