Show that $\left(1+\dfrac{1}{n}\right)^n$ is monotonically increasing

We use the inequality between the geometric mean and the arithmetic mean for the following positive numbers $$ x_{1}=1,~x_{2}=x_{3}=\ldots=x_{n+1}=1+\frac{1}{n}\text{.}% $$ Then $$ \sqrt[n+1]{x_{1}x_{2}\cdots x_{n+1}}<\frac{x_{1}+x_{2}+\ldots+x_{n+1}}{n+1}% $$ (the inequality is strict, since the numbers can't be all equal) translates to $$ \left( 1+\frac{1}{n}\right) ^{\frac{n}{n+1}}<\frac{1+n\left( 1+\frac{1}{n}\right) }{n+1}=1+\frac{1}{n+1}% $$ hence $a_{n}<a_{n+1}$.

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

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

The (well known) elementary proof that this sequence is increasing relies on the Bernoulli inequality, which states that, for real $x\ge -1$ and $n\in \mathbb{N}$, $$(1+x)^n \ge 1+nx$$ which can be easily shown by induction. This looks quite inefficient but should not be underestimated. If you know this, then observe that $$\left(1+\frac{1}{n}\right)^{n} > \left(1+\frac{1}{n-1}\right)^{n-1} $$ is equivalent to $$\left( \frac{1+\frac{1}{n}}{1+\frac{1}{n-1}}\right)^{n} > \left(1+\frac{1}{n-1}\right)^{-1} = 1-\frac{1}{n}$$ The lhs is equal to $$ \left(\frac{n^2-1}{n^2}\right)^n = \left(1-\frac{1}{n^2}\right)^n $$ which, according to Bernoulli is $$> 1-\frac{n}{n^2} = 1-\frac{1}{n}$$ which is what was to be shown.