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

Let $a_n=\left(1+\frac1n\right)^n$.  We need to show first that $\lim_{n\to \infty}a_n$ actually exists.  
From the OP, we see that $a_n$ is bounded above by the number $e$, which is defined as $1=\int_1^e \frac1t\,dt$.  And in THIS ANSWER, I showed using Bernoulli's Inequality that $a_n$ is monotonically increasing.  Inasmuch as $a_n$ is monotonically increasing and bounded above, then $\lim_{n\to \infty}a_n$ does indeed exist.
Next, we see from the OP that 
$$a_n\le e\le a_{n-1}\left(1+\frac1{n-1}\right) \tag 1$$
Since we have established convergence of $a_n$, simply applying the squeeze theorem to $(1)$ yields the coveted equality
$$\lim_{n\to \infty}a_n=e$$
And we are done!
Note that we used $\lim_{n\to \infty}a_{n-1}=\lim_{n\to \infty}a_n$ along with $\lim_{n\to \infty}\left(1+\frac{1}{n-1}\right)=1$ on the right-hand side of $(1)$ in applying the squeeze theorem.
