$$ a_n =\left(1+\frac{1^2}{n^2}\right)^1\left(1+\frac{2^2}{n^2}\right)^2\cdots \left( 1 + \frac{n^2}{n^2} \right)^n$$
$$\lim_{n\to\infty} a_n^{-1/n^2}$$
So I tried solving it by taking the logarithm.
Let the limit be $L$.
Hence,
$$\lim_{n\to\infty}\log(L) = \lim_{n\to\infty} \left(-\frac{1}{n^2}\log\left(1+\frac{1}{n^2}\right)-\frac{2}{n^2}\log\left(1+\frac{2^2}{n^2}\right)-\cdots - \frac{n}{n^2}\log\left(1+\frac{n^2}{n^2}\right)\right).$$
This looks like it should be tractable using the definition of
$$\int^1_0 f(x)\,dx =\lim_{n \to \infty} \frac{1}{n}\sum^n_{r=0}f\left(\frac{r}{n}\right)$$
with taking $f(x) = \log(1+x^2)$ but I am not being able to simplify it to the requisite form. I took this approach mainly because we were recently taught this in school, and it seems to work quite well. I would be interested in alternative solutions too of course. Thanks in advance.