limits - Is $lim_{n to infty} frac{n}{(n!)^frac{1}{n}} = e$ any easier than Stirling?

Stirling's approximation says that
$$\lim_{n \to \infty} \frac{n^n \sqrt{n}}{n! e^n } = \frac{1}{\sqrt{2 \pi}}.$$

Since $\lim_{n \to \infty} x^\frac{1}{n} \to 1$ uniformly on a neighbourhood of $\frac{1}{\sqrt{ 2 \pi}}$, it follows that
$$\lim_{n \to \infty} \left( \frac{n^n \sqrt{n}}{n! e^n } \right)^\frac{1}{n} = \lim_{n \to \infty} \frac{n}{(n!)^\frac{1}{n}} \cdot \frac{n^\frac{1}{2n}}{e} = 1.$$
Since $\lim_{n \to \infty} n^\frac{1}{2n} = 1$, we get the limit in the title
$$\lim_{n \to \infty} \frac{n}{(n!)^\frac{1}{n}} = e.$$

Question: Is Stirling's approximation is really needed to derive the above limit? Or is there an easier way to reach the same conclusion?

Motivation: The radius of convergence $R$ of a power series $\sum_{n=0}^\infty a_nx^n$ is given by Hadamard's formula

$$\frac{1}{R} = \limsup_{n \to \infty} |a_n|^\frac{1}{n}.$$
If we know ahead of time that $R > 0$ then the coefficients are given by
$$a_n = \frac{f^{(n)}(0)}{n!}$$
where $f$ is the function defined by the power series. Then we get
$$\frac{1}{R} = \limsup_{n \to \infty} |a_n|^\frac{1}{n} = \limsup_{n \to \infty} \frac{|f^{(n)}(0)|^\frac{1}{n}}{n} \cdot \frac{n}{(n!)^\frac{1}{n}} = e \cdot \limsup_{n \to \infty} \frac{|f^{(n)}(0)|^\frac{1}{n}}{n}.$$
So one use of the limit is to clean up the formula for the radius of convergence of a power series in terms of the derivatives of the corresponding function.

Answer

HINT: write $\lim_{n \to \infty} \displaystyle\frac{n}{(n!)^\frac{1}{n}}$ as $\lim_{n \to \infty} \left(\displaystyle\frac{n^n}{(n!)}\right)^\frac{1}{n}$ and then compute $\lim_{n \to \infty} \frac{a_{n+1}}{a_{n}}$ where $a_n=\displaystyle\frac{n^n}{(n!)}$