analysis - Evaluating $lim_{n to infty}frac{1}{n}(n!)^{frac{1}{n}}$

I'm trying to find and prove the value of$$\lim_{n \to \infty}\frac{1}{n}(n!)^{\frac{1}{n}}$$

I was thinking that since
$$\frac{1}{n}(n!)^{\frac{1}{n}} = \frac{1}{n} \left[ (1)^{\frac{1}{n}}(2)^{\frac{1}{n}}...(n)^{\frac{1}{n}} \right]$$

and we know that
$$\lim_{n\to \infty}n^{\frac{1}{n}} = 1$$
and
$$k^{\frac{1}{n}} \leq \ n^{\frac{1}{n}} \ \ \ \ \forall k \leq n$$

So $(n!)^{\frac{1}{n}} \leq 1 \ \forall n \ $* Then it is bounded. We also know that $\lim \frac{1}{n} =0$, therefore
$$\lim_{n \to \infty}\frac{1}{n}(n!)^{\frac{1}{n}} = 0 **$$

I'm pretty sure this line of reasoning is ok. Now proving it is another thing. Any suggestions?

*I realize now that this statement is not true, but that did not get me any closer to solving the problem. I do believe that it is bounded though.

**LOL, I don't believe this to be true anymore either, a wild guess tells me that the solution may be $\frac{1}{e}$