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.

We show that
$$
\left(1+\frac{1}{n-1}\right)^{n-1} \leq \left(1+\frac{1}{n}\right)^{n}
$$
for any $n \geq 2$.
For any $n \geq 2$ we have that
$$
n \int^n_{n-1} \frac{1}{x(x+1)}dx \leq \int^n_{n-1} \frac{1}{x}dx
$$
because $\frac{n}{x+1} \leq 1$ for all $x \in [n-1,n]$.
But this inequality is equivalent to
$$
n \int^n_{n-1} \frac{1}{x}-\frac{1}{x+1}dx \leq \int^n_{n-1} \frac{1}{x}dx \iff\\
\iff (n-1)\int^n_{n-1} \frac{1}{x}dx \leq n\int^n_{n-1} \frac{1}{x+1}dx \iff\\
\iff (n-1)\int^n_{n-1} \frac{1}{x}dx \leq n\int^{n+1}_{n} \frac{1}{x}dx \iff\\
\iff (n-1)\ln\left(\frac{n}{n-1}\right)  \leq n \ln\left(\frac{n+1}{n}\right) \iff\\
\iff \left(1+\frac{1}{n-1}\right)^{n-1} \leq \left(1+\frac{1}{n}\right)^{n}.
$$
