Prove $\left(1+\frac1{n}\right)^{n}<\left(1+\frac1{n+1}\right)^{n+1}$ by mathematical induction I am trying to prove the following by mathematical induction:
$$\left(1+\frac1{n}\right)^{n}<\left(1+\frac1{n+1}\right)^{n+1}$$
Other proofs without induction are found here: I have to show $(1+\frac1n)^n$ is monotonically increasing sequence.
But I am curious whether it can be proved by induction as well.
What I've tried so far:
The original inequality is equivalent to
$$(n+1)^{2n+1}<n^n(n+2)^{n+1}$$
So I have to show:
$$(n+2)^{2n+3}<(n+1)^{n+1}(n+3)^{n+2}$$
And,
$$(n+1)^{n+1}(n+3)^{n+2}=(n+1)\color{red}{n^n}\left(1+\frac1n\right)^n\cdot(n+3)\color{red}{(n+2)^{n+1}}\left(1+\frac1{n+2}\right)^{n+1}$$$$>(n+1)(n+3)\cdot\color{red}{(n+1)^{2n+1}}\left(1+\frac1n\right)^n\left(1+\frac1{n+2}\right)^{n+1}$$$$=(n+1)(n+3)\left(n+2+\frac1n\right)^n\left(n+1+\frac{n+1}{n+2}\right)^{n+1}$$
and I am stuck.
 A: Let us define
\begin{align}
t_n :=\left(1+\frac{1}{n} \right)^n.
\end{align}
We shall use the identity
\begin{align}
t_n =&\ 1+1+\frac{1}{2!}\left(1-\frac{1}{n}\right)+\frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)+\ldots\\
&\ +\frac{1}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{n-1}{n}\right).
\end{align}
Inductive Step: Assume it holds for $n=k$, i.e.  $t_{k-1}\leq t_k$. Then observe
\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)\cdots\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)\cdots\left(1-\frac{k}{k+1}\right)\\
\geq&\ 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)\cdots\left(1-\frac{k-1}{k}\right)=t_k. 
\end{align}
Hence the inductive step holds. 
Note: We didn't use the inductive hypothesis.
