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.

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.
