Evaluation of $\lim_{n\rightarrow \infty}\frac{n!\cdot e^n}{\sqrt{n}n^n}$ without stirling approximation

Sorry, your question is not clear. What do you mean "without Stirling approximation" ? Because you use Stirling's construction.

$\displaystyle c:=\lim_{n\to\infty}\frac{n!e^n}{\sqrt{n} n^n}$

$\displaystyle =>\enspace \lim_{n\to\infty}\frac{(2n)!e^{2n}}{\sqrt{2n} (2n)^{2n}}=c ~~, ~~~\lim_{n\to\infty}\frac{n!^2 e^{2n}}{\sqrt{n}^2 n^{2n}}=c^2$

We devide the second by the first, means $c^2/c$ :

$\displaystyle =>\enspace c=\sqrt{2}\lim_{n\to\infty}\frac{n!^2 2^{2n} }{ \sqrt{n} (2n)! }=\sqrt{2}~\Gamma\left(\frac{1}{2}\right)=\sqrt{2\pi} $

If the Gamma function is not enough then look at the Wallis product to get the value of $~c~$ .

Let $$ a_n : = \frac{{n!e^n }}{{n^n \sqrt n }} $$ and $a:= \lim a_n$. I assume that this limit exists and is finite and positive. Then \begin{align*} a_{2n} & = \frac{{(2n)!e^{2n} }}{{(2n)^{2n} \sqrt {2n} }} = \frac{{(2n)!}}{{n!^2 2^{2n} }}\frac{{(n!)^2 e^{2n} }}{{n^{2n + 1} }}\sqrt {\frac{n}{2}} = \sqrt {\frac{{1 \cdot 1}}{{2 \cdot 2}}\frac{{3 \cdot 3}}{{4 \cdot 4}} \cdots \frac{{(2n - 1)(2n - 1)}}{{2n \cdot 2n}}} a_n^2 \sqrt {\frac{n}{2}} \\ & = \sqrt {\frac{{1 \cdot 3}}{{2 \cdot 2}}\frac{{3 \cdot 5}}{{4 \cdot 4}} \cdots \frac{{(2n - 1)(2n + 1)}}{{2n \cdot 2n}}} a_n^2 \sqrt {\frac{n}{{4n + 2}}} . \end{align*} Taking the limit of both sides gives $$ a = \sqrt {\mathop {\lim }\limits_{n \to + \infty } \frac{{1 \cdot 3}}{{2 \cdot 2}}\frac{{3 \cdot 5}}{{4 \cdot 4}} \cdots \frac{{(2n - 1)(2n + 1)}}{{2n \cdot 2n}}} a^2 \frac{1}{2}. $$ The infinite product under the square root is the reciprocal of the famous Wallis product, whence $$ a = \sqrt {\frac{2}{\pi }} a^2 \frac{1}{2}, $$ i.e., $a=\sqrt{2\pi}$.