As is well known, $f(x)=(1+\frac{1}{x})^x$ converges to $e$ as $x\rightarrow\infty$. Sometimes it is good to know that this convergence is monotone which for future reference I will discuss here. We first note that $f(x)=\exp(g(x))$ where
$$\label{1}\tag{1}
g(x)=x\ln\left(1+\frac{1}{x}\right)
$$ and $\exp(x)=e^x$. So it suffices to show $g$ is monotonically increasing to the limit $1$. To that end, rewrite
$$
g(x)=\frac{\ln(1+\frac1x)}{1/x}
$$
and then apply L’Hopital’s rule, and we have that the limit is 1.
How do we know the convergence is monotone? Well, using (\ref{1}) it is readily calculated that $g”(x)<0$ for all $x>0$.

Claim: A function that is concave down and converges to a finite limit as $x\rightarrow\infty$ must necessarily be increasing.

Proof: If we ever had $g’(a)<0$ then $g’(x)\le g’(a)<0$ for all $x\ge a$. By the Mean Value Theorem, on every interval $[n,n+1]$ with $n\ge a$ the function $g$ decreases by at least $|g’(a)|$.

(Namely, if $g(n+1)-g(n)> g’(a)$ then ${\frac{g(n+1)-g(n)}{(n+1)-n}} > g’(a)
$ and so there would be some $x\in [n,n+1]$ with $g’(x)>g’(a)$.)

Consequently $\lim_{x\rightarrow\infty}g(x)=-\infty$.
End of proof of claim.