Simplest or nicest proof that $1+x \le e^x$ The elementary but very useful inequality that $1+x \le e^x$ for all real $x$ has a number of different proofs, some of which can be found online. But is there a particularly slick, intuitive or canonical proof? I would ideally like a proof which fits into a few lines, is accessible to students with limited calculus experience, and does not involve too much analysis of different cases.

Repeatedly using $1 + x \le \left(1 + \frac{x}{2} \right)^2$, we have
\begin{align}
1 + x
\le
\left(1 + \frac x 2\right)^2
\le
\left(1 + \frac x 4\right)^4
\le
\left(1 + \frac x 8\right)^8
\le
\dots
\le
\left(1 + \frac x {2^k}\right)^{2^k}.
\end{align}
Taking the limit of $k \rightarrow \infty$ yields
$$
1 + x \le e^x. \qquad\qquad(1)
$$

Another proof using the technique in this post.  By the AM-GM inequality,
$$
\sqrt[n]{1 \times \cdots \times 1 \times (1 + x)}
\le
\frac{1 + \dots + 1 + (1 + x)}{n}
=1 + \frac{x}{n}.
$$
So,
$$
1+x \le \left(1 + \frac{x}{n} \right)^n.
$$
Taking the limit of $n \rightarrow \infty$ yields (1).
