Why is $\left ( 1+\frac{1}{k} \right )^kIn Artin's Gamma function (page 20), he says in footnote

(..) we consider the elementary inequalities
$$
\left ( 1+\frac{1}{k} \right )^k<e<\left ( 1+\frac{1}{k} \right )^{k+1}
$$
for $k=1,2,\dots, n-1$.

Where does the inequalities come from? From what I know is that
$$
e=\lim_{n\to\infty}\left ( 1+\frac{1}{n} \right )^n.
$$
Is it because that $n\mapsto (1+1/n)^n$ is increasing?
 A: Partial answer.
Let $t_k = \left( 1 + \frac{1}{k} \right) ^k.$
"Borrowing" from Rudin's PMA, in Theorem $3.31$ he gives the nice formula, due to the Binomial theorem:
$$t_k = 1 + 1 + \frac{1}{2!}\left( 1 - \frac{1}{k} \right) + \frac{1}{3!}\left( 1 - \frac{1}{k} \right) \left( 1 - \frac{2}{k} \right) + \ldots$$
$$ + \frac{1}{k!}\left( 1 - \frac{1}{k} \right) \left( 1 - \frac{2}{k} \right) \ldots \left( 1 - \frac{k-1}{k} \right)$$
Hence,
\begin{align} t_{k+1} = 1 + 1 + \frac{1}{2!}\left( 1 - \frac{1}{k+1} \right) + \frac{1}{3!}\left( 1 - \frac{1}{k+1} \right) \left( 1 - \frac{2}{k+1} \right) + \ldots\\
\\
\\ + \frac{1}{k!}\left( 1 - \frac{1}{k+1} \right) \left( 1 - \frac{2}{k+1} \right) \ldots \left( 1 - \frac{k-1}{k+1} \right) \\
\\
\\+ \frac{1}{(k+1)!}\left( 1 - \frac{1}{k+1} \right) \left( 1 - \frac{2}{k+1} \right) \ldots \left( 1 - \frac{k}{k+1} \right)\\
\\
\\ > 1 + 1 + \frac{1}{2!}\left( 1 - \frac{1}{k+1} \right) + \frac{1}{3!}\left( 1 - \frac{1}{k+1} \right) \left( 1 - \frac{2}{k+1} \right) + \ldots \\
\\
\\ + \frac{1}{k!}\left( 1 - \frac{1}{k+1} \right) \left( 1 - \frac{2}{k+1} \right) \ldots \left( 1 - \frac{k-1}{k+1} \right)\\
\\
\\ > t_k.\\
\end{align}
This proves that $\ t_k\ $ is strictly increasing. Since we know that $
\displaystyle\lim_{k\to\infty}\left( 1+\frac{1}{k} \right )^k=e,$ we have the left half of the inequality.
I want to prove the right half of the inequality in the same way, although this is trickier, I think. I'll try again later.
A: We know the following limit$$e = \lim_{k\to \infty}\left(1+\frac 1k\right)^k$$
Now, $f(k) = \left(1+\frac 1k\right)^k \equiv (1 + x)^u \equiv 1+ ux + \frac {u(u-1)x^2}{2!} + \frac {u(u-1)(u-2)x^3}{3!}...+$
We can expand the above function $f(k\to\infty)$ as $\frac 1k$ becomes very small
$$\begin{align*}
\lim_{k\to \infty}f(k)
& = \lim_{k\to \infty}\left[ 1+ \frac kk + \frac {k(k-1)}{2!}\frac 1{k^2} + \frac {k(k-1)(k-2)}{3!}\frac 1{k^3}...+\right]\\
& = \left[ 1+ \lim_{k\to \infty}\frac kk + \lim_{k\to \infty}\frac {k(k-1)}{2!}\frac 1{k^2} + \lim_{k\to \infty}\frac {k(k-1)(k-2)}{3!}\frac 1{k^3}...+\right]\\
& = \left[1+ \frac 11 + \frac 1{2!} + \frac 1{3!}...+\right] = e\\
\end{align*}$$

*

*Theory: When the value of $k\to \infty$ then $f(k) \to e$ but as the derivative of $f(k)$ is positive the function$f(k)$ in increasing function for $k >0$ means if $k$ is some what less than infinity (which for sure no value is equal to infinity)

*Conclusion The above implies that $f(k)$ tends to $e$ but is never greater than $e$ but instead is some what less the $e$ always  thus we can say $f(k) < e$
Now, $F(k) = (1+\frac 1k)^{k+1} =(1+\frac 1k)^k\times(1+\frac 1k) = f(k)(1+\frac1k) = f(k)\times(1+\delta)$
$F(k\to\infty) = f(k\to\infty)(1+\delta) = e \times(\text{something which is greater than 1}) > e$
$e< F(k)$
Thus, $f(k) < e< F(k) = \left(1+\frac1k\right)^k < e < \left(1+ \frac 1k\right)^{k+1}$
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