Is this approach to the limit solution kosher?$ \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\frac{1}{e}$ Here is a non-rigourous approach to a limit:
$$ \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\frac{1}{e}$$
I know the solution is correct. I'm suspicious of my methods.

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*Rewrite the expression:

$$ \lim_{n\rightarrow\infty}\left(\frac{n!}{n^n}\right)^{\frac{1}{n}}$$


*I know the $\lim_{n\rightarrow\infty}\left(\frac{n!}{n^n}\right)=0$. So bear with me.

$$\frac{n!}{n^n}=\frac{n(n-1)(n-2)...[n-(n-2)][n-(n-1)]}{n^n}$$


*Written this way, the numerator is a polynomial of degree $n$. For large $n$, the leading term will prevail. I will write that polynomial as a sum of $n^n$ and a polynomial of degree $n-1$.

$$\frac{n!}{n^n}\Rightarrow\frac{n^n+[\text{polynomial}]^{n-1}}{n^n}\Rightarrow 1+\frac{1}{n}$$


*As noted in Step 2, this is not the expected result for this limit. But watch:

$$ \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{\tfrac{1}{n}}=\lim_{n\rightarrow\infty}\frac{1}{\left(1+\frac{1}{n}\right)^n}=\frac{1}{e}$$
This is the correct solution. But Step 2 is sketchy. Did I make another error which undid this error? Can this derivation be salvaged, or is my proof "not even wrong"?
Please be gentle with me.

This can be considered an heuristic way to proceed, to be rigorous we can use the ratio-root criterion as explained here

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*Why does $\;\lim_{n\to \infty }\frac{n}{n!^{1/n}}=e$?

As noticed there is a mistake in your heuristic derivation, we could try to proceed as follows
$$\left(\frac{n!}{n^n}\right)^n =1 \cdot \left(1-\frac1n\right)^n\cdot \ldots \cdot\left(1-\frac{n-1}n\right)^n \sim \frac1{e^0}\cdot\frac1{e^1}\cdot \ldots \cdot\frac1{e^{n-1}}=\frac1{e^{\frac{n(n-1)}2}}$$
but we would conclude that
$$\left(\frac{n!}{n^n}\right)^\frac1n \sim \frac1{e^{\frac{n(n-1)}{2n^2}}} \implies \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\frac1{\sqrt e}$$
