limits - Show $limleft ( 1+ frac{1}{n} right )^n = e$ if $e$ is defined by $int_1^e frac{1}{x} dx = 1$

I have managed to construct the following bound for $e$, which is defined as the unique positive number such that $\int_1^e \frac{dx}x = 1$.

$$\left ( 1+\frac{1}{n} \right )^n \leq e \leq \left (\frac{n}{n-1} \right )^n$$

From here, there must surely be a way to deduce the well-known equality

$$\lim_{n \rightarrow \infty} \left ( 1+ \frac{1}{n} \right )^n = e$$

I have come up with the following, but I am not absolutely certain if this is correct or not.

PROPOSED SOLUTION:

The lower bound is fine as it is, so we shall leave it alone. Note that

$$\begin{align*} \left ( \frac{n}{n-1} \right )^n &= \left ( 1+\frac{1}{n-1} \right )^{n} \\ &= \left ( 1+\frac{1}{n-1} \right )^{n-1} \left ( 1+\frac{1}{n-1} \right ) \end{align*}$$

So using the fact that the limit distributes over multiplication, we have

$$\lim_{n \rightarrow \infty} \left ( \frac{n}{n-1} \right )^n = \lim_{n \rightarrow \infty} \left ( 1+\frac{1}{n-1} \right )^{n-1} \lim_{n \rightarrow \infty} \left ( 1+\frac{1}{n-1} \right )$$

Since

$$\lim_{n \rightarrow \infty} \left ( 1+\frac{1}{n-1} \right ) = 1$$

and

$$\lim_{n \rightarrow \infty} \left ( 1+\frac{1}{n-1} \right )^{n-1} = \lim_{m \rightarrow \infty} \left ( 1+\frac{1}{m} \right )^m = e$$

We then have the required result

$$\lim_{n \rightarrow \infty} \left ( 1+ \frac{1}{n} \right )^n = e$$

Answer

The proof as stated is circular, but it's easily fixed, more or less by just rephrasing things.

Problem: When you say "Since ... $\lim\left(1+\frac1m\right)^m=e$... we have the required result" you certainly appear to be assuming what you're trying to prove.

Fix: The original inequality shows that

$$e\ge\limsup\left(1+\frac1n\right)^n.$$ Your manipulations with the original upper bound show that $$e\le\liminf\left(1+\frac1n\right)^n,$$ and these two inequalities show that the limit in question exists and equals $e$.