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 $$\lim_{x\rightarrow\infty}\frac{e^x-1}{x}=\infty \quad \text{and} \quad \lim_{x\rightarrow\infty}\frac{\ln{(1+x)}}{x}=0.$$

Let 
$$a=\lim_{n \to \infty} \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n} \quad ;\quad b=\lim_{n \to \infty} \left(1-\frac 1n\right)^{n^2}{e^n}$$
We've;
\begin{align}
c= \frac ab
=\frac{\displaystyle\lim_{n \to \infty} \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n} }{\displaystyle\lim_{n \to \infty} \left(1-\frac 1n\right)^{n^2} {e^n} }
=\lim_{n \to \infty}\frac{\displaystyle \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n} }{\displaystyle \left(1-\frac 1n\right)^{n^2} {e^n} }
&=\lim_{n \to \infty}\left(\frac {n+1}{n-1}\right)^{n^2} \frac{1}{e^{2n}}
\end{align}
Now, let $n=2m$. Since $n \to \infty$, $m \to \infty$ too.
\begin{align}
c&= \lim_{n \to \infty}\left(\frac {n+1}{n-1}\right)^{n^2} \frac{1}{e^{2n}}\\
&=\lim_{m \to \infty}\left(\frac {2m+1}{2m-1}\right)^{(2m)^2} \frac{1}{e^{2(2m)}}\\
&=\lim_{m \to \infty}\left(1+\frac {2}{2m-1}\right)^{4m^2} \frac{1}{e^{4m}}\\
\end{align}
Let $m-\frac 12=p$.
\begin{align}
c&=\lim_{m \to \infty}\left(1+\frac {2}{2m-1}\right)^{4m^2} \frac{1}{e^{4m}}\\
&=\lim_{p \to \infty}\left(1+\frac {1}{p}\right)^{4p^2+4p+1} \frac{1}{e^{4p+2}}\\
&=\lim_{p \to \infty}\left(1+\frac {1}{p}\right)^{4p^2} \left(1+\frac {1}{p}\right)^{4p}\left(1+\frac {1}{p}\right)\frac{1}{e^{4p} \cdot e^2}\\
&=\left(\left(1+\frac {1}{p}\right)^{p^2}\frac {1}{e^p}\right)^4 \left(1+\frac 1p\right)^{4p} \frac{1}{e^2}\\
&=a^4 \cdot e^4 \cdot \frac{1}{e^2}\\
&=a^4 e^2\\
\end{align}
Thus we have $\dfrac ab=a^4 e^2$ $$ \color{blue}{\implies a^3b=\frac{1}{e^2}} \tag 1$$
Now, we also have
\begin{align}
ab=\lim_{n \to \infty} \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n} \cdot \lim_{n \to \infty} \left(1-\frac 1n\right)^{n^2}{e^n}&=\lim_{n \to \infty} \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n}\cdot \left(1-\frac 1n\right)^{n^2}{e^n}\\
&=\lim_{n \to \infty}\left(1-\frac {1}{n^2}\right)^{n^2}\\
&=\frac{1}{e}
\end{align}
Hence,
$$\color{red}{\implies ab=\frac{1}{e}} \tag 2$$
Using $(1)$ and $(2)$ we finally have 
$$\bbox[5px,border:2px solid #6b2fed]{a=\lim_{n \to \infty} \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n} =\frac{1}{\sqrt e}}$$
