<s>[INST] 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.
 [/INST] One that's not been mentioned so far(?): knowing that
$$
0 < e^x = 
 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots
$$
proves the inequality except for $-1 < x < 0$.  But in that region
$$
e^x - (1+x) = 
 \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots
$$
is an alternating series whose terms decrease in absolute value
and start out positive.  Therefore it is positive by the usual argument:
group the terms as
$$
e^x - (1+x) =
\left( \frac{x^2}{2} + \frac{x^3}{3!} \right) + 
\left( \frac{x^4}{4!} + \frac{x^5}{5!} \right) + \cdots
$$
and observe that each combined term is positive, QED.
(This actually works for $-3 < x < 0$, but you still want to use
$e^x > 0$ to prove the inequality for very negative $x$.)
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