How to show $(1/n!)^{1/n}$ goes to $0$ as $n$ goes to infinity? [duplicate]

How do I show $(1/n!)^{1/n}$ goes to $0$ as $n$ goes to infinity? I need this to use the spectral radius theorem to show an operator has spectrum {0}.

Using the AGM-inequality and looking at the graph of $x\mapsto {1\over x}$ we see that $$0<a_n:=\left({1\over n!}\right)^{1/n}\leq {1\over n}\sum_{k=1}^n{1\over k}\leq {1\over n}\left(1+\int_1^n{1\over t}\ dt\right)={1+\log n\over n}\qquad(n\geq 1)\ .$$ It follows that $\lim_{n\to\infty} a_n=0$.

The function $e^x$ is an entire function and hence the Taylor series of $$e^x = 1 + x + \dfrac{x^2}{2!} + \cdots + \dfrac{x^n}{n!} + \cdots $$ has radius of convergence as $\infty$. Hence, $$\lim_{n \to \infty} \left(\dfrac{x^n}{n!} \right)^{1/n} < 1, \,\,\,\,\, \forall x \in \mathbb{R} \implies \lim_{n \to \infty} \left(\dfrac{x}{n!^{1/n}} \right) < 1, \,\,\,\,\, \forall x \in \mathbb{R}$$ Hence, $$\lim_{n \to \infty} \dfrac1{(n!)^{1/n}} = 0$$