Prove by induction... With exponent and factorial

Question

Could someone please help me to prove this by induction?

$$ \left(1+\frac{1}{1}\right) ^1\cdot \left( 1+\frac{1}{2}\right)^2\cdot...\cdot\left( 1+\frac{1}{n-1}\right)^{n-1}=\frac{n^n}{n!}, $$ for $n\ge2$

Thanks!

Answer 1

$n\geq 2$ is the domain of $n$. Then start with the base case

$\texttt{Base case:}$ For $n=2$ we have $\left(1+\frac{1}{1}\right)=\frac{2^2}{2}\Rightarrow 2=\frac42 \checkmark$

$\texttt{Inductive step:}$ Let $X= \left(1+\frac{1}{1}\right) ^1\cdot \left( 1+\frac{1}{2}\right)^2\cdot...\cdot\left( 1+\frac{1}{n-1}\right)^{n-1}$

Then for $n+1$ the equation is

$\left( 1+\frac{1}{n}\right)^{n}\cdot X=\frac{(1+n)^{n+1}}{ (n+1)!}$

$\left( 1+\frac{1}{n}\right)^{n}\cdot X=\frac{(1+n)^n\cdot (n+1)}{n!\cdot (n+1)}$

Canceling out $(n+1)$.

$\left( 1+\frac{1}{n}\right)^{n}\cdot X=\frac{(1+n)^n\cdot }{n!}$

We have $n\cdot \left ( 1+\frac{1}{n}\right)=n+1$. Therefore $\left( 1+\frac{1}{n}\right)^{n}=\left(\frac{n+1}{n}\right)^n $

I think you can take it frome here.