Define the function
\[
f(x):=\left( 1+\frac{1}{x}\right)^x \quad (x>0).
\]
We investigate some properties of it.
1) Limit at $0$.
$\displaystyle{\lim_{x\to 0+}f(x)=\exp\left[\lim_{x\to 0+}x\ln\left(1+\frac{1}{x} \right)\right]}$.
Here
\[
\lim_{x\to 0+}x\ln\left(1+\frac{1}{x}\right)=\lim_{t\to\infty} \frac{\ln(1+t)}{t}=0,
\]
so we get
\[
\lim_{x\to 0+}f(x)=1.\qquad\blacksquare
\]
2) Limit at $\infty$.
$\displaystyle{\lim_{x\to\infty}f(x)=\exp\left[\lim_{x\to\infty}x\ln\left(1+\frac{1}{x} \right)\right]}$.
Here
\[ \lim_{x\to\infty}x\ln\left(1+\frac{1}{x}\right)=\lim_{t\to 0+} \frac{\ln(1+t)}{t}=1, \] so we get \[ \lim_{x\to\infty}f(x)=e.\qquad\blacksquare \]
3) Monotonicity.
\[ f'(x)=\left[\left(1+\frac{1}{x} \right)^x\right]'=\left(1+\frac{1}{x} \right)^x\left[\ln\left(1+\frac{1}{x} \right)-\frac{1}{x+1} \right]. \]
Take
\[
g(x):=\ln\left(1+\frac{1}{x} \right)-\frac{1}{x+1}.
\]
Then
\[
g'(x)=-\frac{1}{x(x+1)^2},
\]
so $g(x)$ is strictly decreasing. Since $\lim_{x\to\infty}g(x)=0$ it implies that $g(x)>0$. Thus it follows $f'(x)>0$ that yields the function $f(x)$ is strictly increasing.$\qquad\blacksquare$
4) Concavity.
\begin{align*} f''(x)=\frac{1}{x(x+1)^2}\left(1+\frac{1}{x}\right)^x &\left[ x(x+1)^2\ln^2\left(1+\frac{1}{x} \right)\right.\\ &\left.-2(x^2+x)\ln\left(1+\frac{1}{x} \right) +x-1 \right]. \end{align*} Take
\[
h(x):=x(x+1)^2\ln^2\left(1+\frac{1}{x} \right)-2(x^2+x)\ln\left(1+\frac{1}{x} \right) +x-1.
\]
We show that $h(x)<0$ for $x>0$.
Assume that there is a positive root of $h(x)$.
Then
\[
\ln\left(1+\frac{1}{x} \right)=\frac{\sqrt{x}+1}{\sqrt{x}(x+1)}
\]
or
\[
\ln\left(1+\frac{1}{x} \right)=\frac{\sqrt{x}-1}{\sqrt{x}(x+1)}.
\]
We have seen in 3) that $\ln\left(1+\frac{1}{x} \right)>\frac{1}{x+1}$, so the second equality is impossible.
Now we consider the first one. Define the function
\[
m(x):=\ln\left(1+\frac{1}{x} \right)-\frac{\sqrt{x}+1}{\sqrt{x}(x+1)}.
\]
We show that $m(x)<0$ for $x>0$ so the equality is impossible.
\[
m'(x)=\frac{-2x^{1/2}+3x+1}{2x^{3/2}(x+1)^2}.
\]
Here the numerator is a quadratic polynomial in $x^{1/2}$, and the discriminant of $-2z+3z^2+1$ is negative. It means $m'(x)$ does not change its sign. Since $\lim_{x\to 0+}m'(z)=\infty$ it implies $m'(x)>0$. Thus we obtain that $m(x)$ is strictly increasing. Since $\lim_{x\to\infty}m(x)=0$ it proves that $m(x)<0$.
Until now we showed that $h(x)$ does not change its sign.
Since $\lim_{x\to 0+}h(x)=-1$ we have justified that $h(x)<0$ for $x>0$. $\blacksquare$
Graph of $\left(1+\frac{1}{x}\right)^x$
