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.

*

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

*

*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
$$\left(\frac{n!}{n^n}\right)^\frac1n \sim \frac1{e^{\frac{n(n-1)}{2n^2}}} \implies \lim_{n\rightarrow\infty}\frac{(n!)^{\tfrac{1}{n}}}{n}=\frac1{\sqrt e}$$
A: In step 2 you don't consider that
$\frac{n^n+[\text{polynomial}]^{n-1}}{n^n}$ has infinite polynomial degree in the limit, so is false
$\frac{n^n+[\text{polynomial}]^{n-1}}{n^n}\rightarrow 1+\frac{1}{n}$.
I recommend using Stirling's formula.
A: Too long for a comment.
Since you already received good answers, I shall try to go beyond the limit. I must precise that, as soon as I see a factorial, I automatically think about Stirling approximation.
$$a_n=\frac{(n!)^{\tfrac{1}{n}}}{n}\implies \log(a_n)=\frac 1n \log(n!)-\log(n)$$
$$\log(n!)=n (\log (n)-1)+\frac{1}{2} \left(\log (2 \pi )+\log
   \left({n}\right)\right)+\frac{1}{12
   n}+O\left(\frac{1}{n^3}\right)$$ which makes
$$\log(a_n)=-1+\frac{\log (2 \pi  n)}{2 n}+O\left(\frac{1}{n^2}\right)$$
$$a_n=e^{\log(a_n)}=\frac 1 e\left(1+\frac{\log (2 \pi  n)}{2 n} +O\left(\frac{1}{n^2}\right)\right)$$
which shows the limit and how it is approached.
Moreover, this gives you a shortcut eveluation of the $a_n$'s. Suppose $n=10$; the exact value is $0.453$ while the above truncated expression gives $0.444$ (that is to say a $2$% relative error).
A: Another very quick way is to use one of the asymptotic Stirling approximations for the factorial function. That is,
$$
n!\sim \frac{n^n}{e^n}
$$
So
$$
\lim_{n\rightarrow\infty}\frac{(n!)^{\frac{1}{n}}}{n}=\lim_{n\rightarrow\infty}\frac{(\frac{n^n}{e^n})^{\frac{1}{n}}}{n}=\lim_{n\rightarrow\infty}\frac{(n)}{e(n)}=\frac{1}{e}
$$
Hence
$$
\lim_{n\rightarrow\infty}\frac{(n!)^{\frac{1}{n}}}{n}=\frac{1}{e}
$$
