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

Suppose
$$
1=\int_1^x\frac{\mathrm{d}u}u\tag{1}
$$
Using the change of variables $u\mapsto u^n$, we get
$$
1=n\int_1^{x^{1/n}}\frac{\mathrm{d}u}u\tag{2}
$$
Dividing by $n$ and estimating the integral using the width of the interval and the extremes of $u$:
$$
1-x^{-1/n}\le\frac1n\le x^{1/n}-1\tag{3}
$$
or equivalently
$$
\left(1+\frac1n\right)^n\le x\le\left(1+\frac1{n-1}\right)^n\tag{4}
$$
Since $x$ is not dependent on $n$, we also have
$$
\left(1+\frac1n\right)^n\le x\le\left(1+\frac1n\right)^{n+1}\tag{5}
$$
In this answer, it is shown that the left side of $(5)$ is an increasing sequence and the right side of $(5)$ is a decreasing sequence. Since the ratio of the right side over the left side is $1+\frac1n$, the left side of $(5)$ increases to $x$ and the right side of $(5)$ decreases to $x$. Therefore,
$$
x=\lim_{n\to\infty}\left(1+\frac1n\right)^n\tag{6}
$$
