sequences and series - Show that $left(1+dfrac{1}{n}right)^n$ is monotonically increasing

Show that $U_n:=\left(1+\dfrac{1}{n}\right)^n$, $n\in\Bbb N$, defines a monotonically increasing sequence.

I must show that $U_{n+1}-U_n\geq0$, i.e. $$\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\geq0.$$

I am trying to go ahead of this step.

Answer

$$x_n=\bigg(1+\frac{1}{n}\bigg)^n\longrightarrow x_{n+1}=\bigg(1+\frac{1}{n+1}\bigg)^{n+1}$$
$$\frac{x_{n+1}}{x_{n}}=\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n})^{n}}=\bigg(\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}\bigg)^n\bigg(1+\frac{1}{n+1}\bigg)=\bigg(\frac{n(n+2)}{(n+1)^2}\bigg)^n\bigg(1+\frac{1}{n+1}\bigg)$$
$$=\bigg(1-\frac{1}{(n+1)^2}\bigg)^n\bigg(1+\frac{1}{n+1}\bigg)≥\bigg(1-\frac{n}{(n+1)^2}\bigg)\bigg(1+\frac{1}{n+1}\bigg)$$
$$≥^*\frac{1}{1+\frac{1}{n+1}}\bigg(1+\frac{1}{n+1}\bigg)≥1$$
It means that your sequence is increasing.

≥*: $$(n+2)(n^2+n+1)=(n+2)\bigg((n+1)^2-n\bigg)≥(n+1)^3$$