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

there are 2 mistakes:
1)if you know $\lim_{n \to \infty} a_n, \lim_{n \to \infty} b_n$ exist, then you know that  $\lim_{n \to \infty} a_n \cdot \lim_{n \to \infty} b_n = \lim_{n \to \infty} a_n\cdot b_n$ exists the other round is false!
2) You may not use the requirement $\lim_{m \to \infty} (1 + \frac{1}{m})^m$ in your proof (logical mistake). 
Try instead natural logarithm over your inequality and that you can pull the limes out of all continous functions $f$ f.e.$$
\lim_{n \to \infty} f(a_n) = f(\lim_{n \to \infty} a_n)$$
For more hints, answer^^
