Using the ratio test on $\sum_{n=1}^{\infty} \left(1+\frac{1}{n}\right)^{n^2}$ I already know that I should do the root test on this series due to the $n$th power, but I want to see if I can establish the result using the ratio test first. (Or would I always be stuck with only one kind of test)
I get something like this: $$\lim_{n \to \infty} \left|\frac{\left(1+\frac{1}{n+1}\right)^{\left(n+1\right)^2}}{\left(1+\frac{1}{n}\right)^{n^2}}\right|$$
But then I'm not sure how (or even if I could) simplify this. I've tried on Wolfram Alpha to make it simplify but the best I can get is numerical approximations that seem to converge to $e$

First, note that
$$
\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}
= 1-\frac{1}{(n+1)^2} \tag{1}
$$
We will use that to make the "right" exponent appear, since we want to use the known limit $\lim_{m\to\infty} (1+\frac{u}{m})^m = e^u$.
Now, since $n^2=(n+1)^2-(2n+1)$, we can rewrite
$$\begin{align*}
\frac{\left(1+\frac{1}{n+1}\right)^{\left(n+1\right)^2}}{\left(1+\frac{1}{n}\right)^{n^2}}
&= \left(1+\frac{1}{n}\right)^{2n+1}\cdot \frac{\left(1+\frac{1}{n+1}\right)^{(n+1)^2}}{\left(1+\frac{1}{n}\right)^{(n+1)^2}}
= \left(1+\frac{1}{n}\right)^{2n+1} \left(\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}\right)^{n^2} \\
&= \color{red}{\left(1+\frac{1}{n}\right)}\cdot \color{blue}{\left(1+\frac{2}{2n}\right)^{2n}}\cdot \color{green}{\left(1-\frac{1}{(n+1)^2}\right)^{(n+1)^2}} \tag{2}
\end{align*}$$
This is great! We know that
$$
\lim_{n\to\infty} \left(1+\frac{1}{n}\right) = \color{red}{1} \tag{3}
$$
and
$$
\lim_{n\to\infty} \left(1+\frac{2}{2n}\right)^{2n} = \color{blue}{e^{2}} \tag{4}
$$
and
$$
\lim_{n\to\infty} \left(1-\frac{1}{(n+1)^2}\right)^{(n+1)^2} = \color{green}{e^{-1}} \tag{5}
$$
so, combining (3), (4), and (5) into (2), we get
$$
\lim_{n\to\infty} \frac{\left(1+\frac{1}{n+1}\right)^{\left(n+1\right)^2}}{\left(1+\frac{1}{n}\right)^{n^2}}
= \color{red}{1} \cdot \color{blue}{e^{2}} \cdot \color{green}{e^{-1}} = \boxed{e}
$$
