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
$$\lim_{n \rightarrow \infty} \left ( 1+ \frac{1}{n} \right )^n = e$$
 A: 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 $e$.
A: David C. Ullrich has already given a perfect answer using the tactical device of $\limsup, \liminf$ which helps a lot when the existence of limit is not known in advance.

However since you have used the integral definition for $e$ it is better to use the properties of integrals to prove your claim. So let $$f(x) = \int_{1}^{x}\frac{dt}{t}\tag{1}$$ for $x > 0$. The integral above does not exist if $x \leq 0$. And the definition of $e$ is given by $f(e) = 1$. We have
\begin{align}
f(xy) &= \int_{1}^{xy}\frac{dt}{t}\notag\\
&= \int_{1}^{x}\frac{dt}{t} + \int_{x}^{xy}\frac{dt}{t}\notag\\
&= f(x) + \int_{1}^{y}\frac{d(vx)}{vx}\text{ (putting }t = vx)\notag\\
&= f(x) + \int_{1}^{y}\frac{dv}{v}\notag\\
&= f(x) + f(y)\notag
\end{align}
Using the above property repeatedly we see that $f(x^{n}) = nf(x)$ for positive integer $n$ and $x > 0$. Therefore if $a_{n} = \left(1 + \dfrac{1}{n}\right)^{n}$ then $$f(a_{n}) = nf\left(1 + \frac{1}{n}\right)\tag{2}$$ We will show that that the RHS of the above equation tends to $1$ and therefore LHS also does the same.
If $1 < t < 1 + 1/n$ then $$\frac{n}{n + 1} < \frac{1}{t} < 1$$ and upon integrating this inequality on interval $[1, 1 + 1/n]$ we get $$\frac{1}{n + 1} < f\left(1 + \frac{1}{n}\right) < \frac{1}{n}$$ or $$\frac{n}{n + 1} < nf\left(1 + \frac{1}{n}\right) < 1$$ and by Squeeze theorem we now see that $$\lim_{n \to \infty}nf\left(1 + \frac{1}{n}\right) = 1$$ and therefore from $(2)$ we get $$\lim_{n \to \infty}f(a_{n}) = 1 = f(e)\tag{3}$$ Now a quick but incorrect route to the answer is to use the continuity of $f$ to replace $\lim f(a_{n})$ by $f(\lim a_{n})$ to get $f(\lim a_{n}) = f(e)$ and then use one-one property of $f$ to get $\lim a_{n} = e$. But the problem with this approach is that we don't know whether limit of $a_{n}$ exists or not (BTW the existence of this limit is routinely proved in many real-analysis/calculus textbooks but we don't use that approach here).
Instead we note that $f(x)$ is differentiable for $x > 0$ with a positive derivative and therefore there exists a unique inverse $g$ of $f$ which is also differentiable with positive derivative. The domain/range of $g$ match the range/domain of $f$ and it can be proved that $f$ maps $\mathbb{R}^{+}$ to $\mathbb{R}$ so that the inverse $g$ maps $\mathbb{R}$ to $\mathbb{R}^{+}$. Applying $g$ on equation $(3)$ we get $$g(\lim_{n \to \infty}f(a_{n})) = g(f(e)) = e$$ and by continuity of $g$ we get $$\lim_{n \to \infty}g(f(a_{n})) = e$$ or $$\lim_{n \to \infty}a_{n} = e$$ which we set out to prove.

It should be obvious to many readers that the functions $f, g$ are $\log$ and $\exp$ respectively, but it is better to use the symbols $f, g$ in order to avoid using any well known properties of $\log, \exp$. This is a great help in avoiding circular proofs.
A: 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.
A: 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,
$$
x=\lim_{n\to\infty}\left(1+\frac1n\right)^n\tag{6}
$$
A: 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^^
A: "$ ( 1+\frac{1}{n}  )^n \leq e \leq  (\frac{n}{n-1})^n$"$
I'm assuming you've shown $( 1+\frac{1}{n}  )^n < ( 1+\frac{1}{n+1}  )^{n+1}$ so we can say $\lim_{n\rightarrow \infty}( 1+\frac{1}{n}  )^n = c \le e$ and likewise $\lim_{n\rightarrow \infty}(\frac{n}{n-1}  )^n = d \ge e$.
So $\frac cd = \lim_{n\rightarrow \infty}\frac{( 1+\frac{1}{n}  )^n}{(\frac{n}{n-1}  )^n}= \lim_{n\rightarrow \infty}(\frac{(n+1)(n-1)}{n^2})^n=$
$\lim_{n\rightarrow \infty}(1- \frac{1}{n^2})^n = \lim 1 - \frac{n}{n^2} = 1$
So $c = d$ and $c = e = d$.
So $c = \lim ( 1+\frac{1}{n}  )^n = e$.
