find the limit of this function? Evaluate $$\lim_{n\to \infty}\frac{1}{(n!)^{1/n}}={?}$$

My try: 
$$\lim_{n\to \infty}\frac{1}{(n!)^{1/n}}=\exp{\lim_{n\to \infty}\frac{-\ln(n!)}{n}}$$
$$=\exp{\lim_{n\to \infty}\frac{-\ln(1\times2\times3\ldots(n-1)\times n)}{n}}$$
$$=\exp{\lim_{n\to \infty}\frac{-(1+2+3+\ldots+(n-1)+n)}{n}}$$
$$=\exp{(-\lim_{n\to \infty}\left(\frac1n+\frac2n+\ldots+\frac{n-1}{n}+\frac nn\right)})$$
$$=\exp(-1)=\frac1e$$
I don't know if my answer is correct please correct me if i am wrong. thanks
Note that $e^x \geq \frac{x^n}{n!}$ for all $x\geq 0$. Setting $x=n$, we have $$ n! \geq \left( \frac{n}{e} \right)^n .$$
Thus $$(n!)^{1/n} \geq \frac{n}{e} \to \infty.$$
Stirling's approximation $$n!\approx\sqrt{2\pi n}\left(\dfrac ne\right)^n$$


$$\begin{align}L&amp;=\lim_\limits{n\to\infty}\dfrac1{(n!)^{\frac1n}}\\&amp;=\lim_\limits{n\to\infty}\dfrac1{(2\pi n)^{\frac 1{2n}}\cdot\dfrac ne}\\&amp;=\lim_\limits{n\to\infty}\dfrac{e}{ne^{\frac{\ln(2\pi n)}{2n}}}\\&amp;=\lim_\limits{n\to\infty}\dfrac e{ne^{\frac1{2n}}}\qquad\left[\text{Applying L' H}\hat{o}\text{pital's Rule on }\lim_\limits{n\to\infty}\dfrac{\ln(2\pi n)}{2n}\right]\\&amp;=\lim_\limits{n\to\infty}\dfrac en\\&amp;=0\end{align}$$