$\lim_{n\rightarrow \infty}\left[\frac{\left(1+\frac{1}{n^2}\right)\cdot \cdots\cdots \left(1+\frac{n}{n^2}\right)}{\sqrt{e}}\right]^n$

We have $$\log{\left ( 1+\frac{r}{n^2} \right )}=\frac{r}{n^2}-\frac12\frac{r^2}{n^4}+O\left(\frac{r^3}{n^6}\right),$$ that means $$n\sum^n_ {r=1}\log{\left ( 1+\frac{r}{n^2} \right )}=\frac{n(n+1)}2\frac1n-\frac12\frac{n(n+1)(2n+1)}6\frac1{n^3}+O(n^4)\frac1{n^5},$$ so we get $$n\sum^n_ {r=1}\log{\left ( 1+\frac{r}{n^2} \right )}-\frac{n}2\to\frac12-\frac16=\frac13$$ as $n\to\infty$.

We have that $$\begin{align} \left[\frac{\left(1+\frac{1}{n^2}\right)\cdots \left(1+\frac{n}{n^2}\right)}{\sqrt{e}}\right]^n&=\frac{\left(\frac{(n^2+n)!}{(n^2)!}\right)^n}{n^{2n^2} e^{n/2}} \sim\frac{\left(1+\frac{1}{n}\right)^{n^3+n^2+\frac{n}{2}}}{e^{n^2+n/2}} \\ &\sim\frac{\exp\left((n^3+n^2+\frac{n}{2})\ln\left(1+\frac{1}{n}\right)\right)}{e^{n^2+n/2}}\\ &\sim\frac{\exp\left((n^2+n+\frac{1}{2})-\frac{n+1}{2}+\frac{1}{3}\right)}{e^{n^2+n/2}}\to e^{1/3} \end{align}$$ where we used the Stirling approximation for factorials and the expansion $$\ln(1+t)= t-\frac{t^2}{2}+\frac{t^3}{3}+o(t^3).$$