How to calculate $\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)$?

We will estimate $$f(n) = \sum_{k=1}^n \log \left(1 + \dfrac{k}{n^2}\right)$$ Recall that $\log(1+x) = x + \mathcal{O}(x^2)$. Hence, we get that $$f(n) = \sum_{k=1}^n \left(\dfrac{k}{n^2} + \mathcal{O}\left(\dfrac{k^2}{n^4}\right)\right) = \dfrac{n+1}{2n} + \mathcal{O}(1/n) \tag{$\star$}$$ where we made use of the fact that $\displaystyle \sum_{k=1}^n k = \dfrac{n(n+1)}2$ and $\displaystyle \sum_{k=1}^n k^2 = \dfrac{n(n+1)(2n+1)}6$.

Now letting $n \to \infty$ in $(\star)$, we get that $$\lim_{n \to \infty} f(n) = \dfrac12$$ The product you are interested in is $e^{f(n)}$ and since $e^x$ is continuous, we have $$\lim_{n \to \infty} e^{f(n)} = e^{\lim_{n \to \infty} f(n)} = e^{1/2} = \sqrt{e}$$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\lim_{n \to \infty}\pars{1 + {1 \over n^{2}}} \pars{1 + {2 \over n^{2}}}\cdots\pars{ 1 + {n \over n^{2}}} = \lim_{n \to \infty}{\pars{n^{2} + 1}\pars{n^{2} + 1}\cdots\pars{n^{2} + n} \over n^{2n}} \\[5mm] = &\ \lim_{n \to \infty}{\pars{n^{2} + 1}^{\large\overline{n}} \over n^{2n}} = \lim_{n \to \infty} {\Gamma\pars{n^{2} + 1 + n}/\Gamma\pars{n^{2} + 1} \over n^{2n}} = \lim_{n \to \infty} {\pars{n^{2} + n}! \over n^{2n}\,\pars{n^{2}}!} \\[5mm] = &\ \lim_{n \to \infty} {\root{2\pi}\pars{n^{2} + n}^{n^{2} + n + 1/2}\expo{-n^{2} - n} \over n^{2n}\bracks{\root{2\pi}\pars{n^{2}}^{n^{2} + 1/2}\expo{-n^{2}}}} = \lim_{n \to \infty} {n^{2n^{2} + 2n + 1}\ \pars{1 + 1/n}^{n^{2} + n + 1/2}\ \expo{-n} \over n^{2n}\pars{n^{2n^{2} + 1}}} \\[5mm] = &\ \exp\pars{\lim_{n \to \infty}\braces{\bracks{n^{2} + n + {1 \over 2}}\ln\pars{1 + {1 \over n}} - n}} \\[5mm] = &\ \exp\pars{\lim_{n \to \infty}\braces{\bracks{n^{2} + n + {1 \over 2}} \bracks{{1 \over n} - {1 \over 2n^{2}}} - n}} \\[5mm] = &\ \exp\pars{\lim_{n \to \infty}\braces{% \bracks{n -{1 \over 2} + 1 - {1 \over 2n} + {1 \over 2n} - {1 \over 4n^{3}}} - n}} = \bbx{\ds{\root{\expo{}}}} \end{align}

For $1\le k\le n$, we have

$$\begin{align} \log\left(1+{k\over n^2}\right)-{k\over n(n+1)} &=\int_1^{1+k/n^2}{dt\over t}-{k\over n(n+1)}\\ &=\int_0^{k/n^2}\left({1\over1+t}-{n\over n+1}\right)dt\\ &={1\over n+1}\int_0^{k/n^2}\left(1-tn\over1+t \right)dt\\ &\le{1\over n+1}\int_0^{k/n^2}dt\\ &\le{k\over n^2(n+1)}\\ &\le{1\over n(n+1)} \end{align}$$

Note, the integral in the third line is clearly non-negative, which means the initial expression is also non-negative. Therefore

$$0\le\log\left(1+{k\over n^2}\right)-{k\over n(n+1)}\le{1\over n(n+1)}$$

Consequently, since $1+2+\cdots+n={n(n+1)\over2}$, we have

$$0\le\sum_{k=1}^n\log\left(1+{k\over n^2}\right)-{1\over 2}=\sum_{k=1}^n\left(\log\left(1+{k\over n^2}\right)-{k\over n(n+1)}\right)\le{1\over n+1}$$

so by the Squeeze Theorem

$$\lim_{n\to\infty}\sum_{k=1}^n\log\left(1+{k\over n^2}\right)={1\over2}$$

and thus

$$\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)=e^{1/2}=\sqrt e$$