<s>[INST] Compute $\lim_{n\rightarrow\infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot\frac{1}{e^n}.$ NOTE 1: L'Hospitals and Taylor expansions are not allowed.
NOTE 2: I really appreciate if someone would correct my attempt, however any other easier method only involving single variable calculus (excluding the concepts in NOTE 1) are welcome.
PROBLEM: Compute $$\lim_{n\rightarrow\infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot\frac{1}{e^n}.$$
I'll just manipulate without writing out the limit, for now. I have 
\begin{array}{lcl}
\left(\frac{n+1}{n}\right)^{n^2}\cdot\frac{1}{e^n} & = & \left( 1+\frac{1}{n}\right)^{n^2}\cdot e^{-n} \\
& = & \exp\left( n^2\ln\left(1+\frac{1}{n}\right)-n\right) \\
& = & \exp((n\ln(1+\frac{1}{n})-1)n) \\
\end{array}
And proceeding:
\begin{array}{lcl}
\exp((n\ln(1+\frac{1}{n})-1)n) & = & e^{((n\ln(1+\frac{1}{n})-1)n} \\
& = & (e^{((n\ln(1+\frac{1}{n})-1)})^n \\
& = & \left(\frac{(e^{((n\ln(1+\frac{1}{n})-1)}-1+1)}{(n\ln(1+\frac{1}{n})-1}\cdot{((n\ln(1+\frac{1}{n})-1})\right)^n\\
\end{array}
It gets quite ugly very quickly as you can see. I'm trying to rewrite it so I can apply standard limits like $$\lim_{x\rightarrow\infty}\frac{e^x-1}{x}=\infty \quad \text{and} \quad \lim_{x\rightarrow\infty}\frac{\ln{(1+x)}}{x}=0.$$
 [/INST] Using the link
"Find the smallest $a>0$ such that $(1+\frac{1}{x})^{x+a}>e$ for all $x\geq 1\,$."
we have $\enspace\displaystyle (1+\frac{1}{n})^n<e<(1+ \frac{1}{n} )^{n+\frac{1}{2}}\enspace$for positive $\,n\,$ and therefore 
$\displaystyle \frac{1}{\sqrt{e}}< (1+\frac{1}{n})^{-\frac{n}{2}}< (1+ \frac{1}{n})^{n^2}e^{-n}<1\enspace$ .
But $\enspace\displaystyle (1+ \frac{1}{x} )^{x^2}e^{-x}\enspace$ is strictly monotonous decreasing for $\,x>0\,$: 
One result in the answer for the question in the link above is $\enspace\displaystyle \ln(1+\frac{1}{x} )<\frac{x+\frac{1}{2}}{x(x+1)}\enspace$ 
with which we get $\enspace\displaystyle \frac{d}{dx}\left(x^2\ln(1+\frac{1}{x}) - x\right) = -\frac{2x+1}{x+1} + 2x\ln(1+\frac{1}{x})) <0 \enspace $ . 
It follows $\enspace\displaystyle \lim\limits_{n\to\infty} (1+ \frac{1}{n})^{n^2}e^{-n} = \frac{1}{\sqrt{e}} \enspace$ .
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