<s>[INST] 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.
 [/INST] Here's a proof I learned by solving problem 90 from Section 11.1 of James Stewart's Calculus: Early Transcendentals (8th edition). We'll need the following result:

Lemma: If $0\leq a<b$ and $n$ is a positive integer, then
$$\frac{b^{n+1}-a^{n+1}}{b-a}<(n+1)b^n$$
Proof: Define the function $f:[0,\infty)\rightarrow\mathbb{R}$ by $f(x)=x^{n+1}$, where $n$ is a positive integer. Then $f$ is differentiable (and hence continuous) over $[0,\infty)$. Moreover, by the power rule, $f'(x)=(n+1)x^n$ for all $x\geq 0$, which is strictly increasing because $n$ is a positive integer (this will be important later, so we'll tuck it in our back pocket).
Since $f$ is differentiable over $[0,\infty)$, and $0\leq a<b$, it is necessarily continuous over $[a,b]$ and differentiable over $(a,b)$, so the mean value theorem asserts that there exists $c\in(a,b)$ such that
$$f'(c)=\frac{f(b)-f(a)}{b-a}
=\frac{b^{n+1}-a^{n+1}}{b-a}$$
Since $f'$ is strictly increasing and $c<b$, it follows that $f'(c)<f'(b)=(n+1)b^n$, so
$$\frac{b^{n+1}-a^{n+1}}{b-a}=f'(c)<f'(b)=(n+1)b^n$$
It immediately follows that $\frac{b^{n+1}-a^{n+1}}{b-a}<(n+1)b^n$.

Simple algebraic manipulation shows that $\frac{b^{n+1}-a^{n+1}}{b-a}<(n+1)b^n$ is equivalent to $b^n[(n+1)a-nb]<a^{n+1}$. With this established, we choose $a=1+\frac{1}{n+1}$ and $b=1+\frac{1}{n}$ (this is justified since $\frac{1}{n+1}<\frac{1}{n}$ for all $n>0$, and hence $1+\frac{1}{n+1}<1+\frac{1}{n}$), which gives
$$\left(1+\frac{1}{n}\right)^n\left[(n+1)\left(1+\frac{1}{n+1}\right)-n\left(1+\frac{1}{n}\right)\right]<\left(1+\frac{1}{n+1}\right)^{n+1}$$
Therefore,
$$\left(1+\frac{1}{n}\right)^n(n+1+1-n-1)<\left(1+\frac{1}{n+1}\right)^{n+1}$$
and thus,
$$\left(1+\frac{1}{n}\right)^n(1)=\left(1+\frac{1}{n}\right)^n<\left(1+\frac{1}{n+1}\right)^{n+1}$$
This shows that $U_n=\left(1+\frac{1}{n}\right)^n$ is strictly increasing.
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