Inequality $ \left(1+\frac{x}{n}\right)^n \leq \exp(x) $ How does one show this seemingly simple, but actually not that simple, inequality

$$
\left(1+\frac{x}{n}\right)^n \leq \exp(x)
$$
In this answer I define the natural logarithm as
$$
\log x =\int_{1}^{x}\frac{1}{t} \, dt \, .
$$
and the exponential function as the inverse of the logarithm.

We begin with the fact that
$$
\frac{x}{n} \geq \log\left(1+\frac{x}{n}\right) \, .
$$
which can be deduced geometrically:

$\log\left(1+\frac{x}{n}\right)$ is the area of the region bounded by the hyperbola $y=1/t$, the $t$-axis, and the vertical lines $t=1$ and $t=1+\frac{x}{n}$. The rectangle with a width of $x/n$ and a height of $1$ gives us an upper bound for this region. A little care is needed when $x/n$ is less than $0$, but the result still holds. Hence,
$$
n\log\left(1+\frac{x}{n}\right) \leq x \, ,
$$
and therefore
$$
\exp\left(n\log\left(1+\frac{x}{n}\right)\right)=\left(1+\frac{x}{n}\right)^n \leq \exp(x) \, .
$$
Assuming you can use the fact that the limit of that sequence is equal to $exp(x)$, then the sequence will be bounded by $exp(x)$, meaning that the inequality will be true.