$a_n=(1+\frac{1}{n})^n$: Prove the following Equation.

Question

I have trouble solving following problem:

Given the sequence:

$$a_n=\Big(1+\frac{1}{n}\Big)^n$$

We have to show that

$$\frac{a_{n+1}}{a_n} = \Big(1-\frac{1}{(n+1)^2}\Big)^{n+1} \frac{n+1}{n} , n \in \mathbb N$$

I would really appreciate any help and I hope everything is correctly written.

Greetings

Answer 1

Because $$\frac{a_{n+1}}{a_n}=\frac{\frac{(n+2)^{n+1}}{(n+1)^{n+1}}}{\frac{(n+1)^n}{n^n}}=\frac{n+1}{n}\cdot\frac{(n+2)^{n+1}n^{n+1}}{(n+1)^{2n+2}}=$$ $$=\frac{n+1}{n}\cdot\left(\frac{n^2+2n}{(n+1)^2}\right)^{n+1}=\left(1-\frac{1}{(n+1)^2}\right)^{n+1}\cdot\frac{n+1}{n}.$$