<s>[INST] 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?
 [/INST] 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.
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