I'm confused on the limit of $\left(1+\frac{1}{n}\right)^n$ Okay so I read Richard Rusczyk's AoPS Volume 2 Book, and I stumbled upon the part where he informs very briefly that $\lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n=e$. But he doesn't really provide a rigorous proof as to why that's true (not criticizing him or anything).. It would really help if someone could provide me with the simplest proof possible as to why $\lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n=e$. Thank you in advance!
Assume $e$ is defined by $e := \sum_\limits{k=0}^{\infty} \frac {1}{k!}$

Let 

$$y=\lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n$$
$$\ln y=\lim_{n \rightarrow \infty} n\ln\left(1+\frac{1}{n}\right)$$
$$\ln y=\lim_{n \rightarrow \infty} \frac{\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}}$$
using L'Hôpital's rule
$$\ln y = \lim_{n \rightarrow \infty} \frac{\frac{-1}{n^{2}}*\frac{1}{1+\frac{1}{n}}}{\frac{-1}{n^{2}}}$$
$$\ln y = \lim_{n \rightarrow \infty} \frac{1}{1+\frac{1}{n}}$$
$$\ln y = 1$$
$$y=e$$
This is of the form $1^\infty$ as $ n \to \infty$.

Now for any limit $\displaystyle\lim_{n\to \infty}f(x)^{g(x)}$ of the form $1^\infty$ we have direct formula to be : $e^{lim_{n\to\infty}(f(x)-1)(g(x))}$.

Now $f(x)=1+\frac{1}{n}$ and $g(x)=n$. Evaluating using the above formula the limit comes out to be : $$e^{lim_{n\to \infty}(1+\frac{1}{n}-1)n}=e^{lim_{n \to \infty}(\frac{1}{n})n}=e.$$