Proof that $A + 1 \leq e^A$ for all $A > 0$ I was reading a proof where at a certain point the prover uses the following inequality
$$A + 1 \leq e^A$$
which in my opinion needs also a proof to be used around. I think I'm missing some important fundamental property which is well-known, but I'm not sure.
 A: In the simple case of $A>0$, it suffices to just look at the taylor series definition:
$$1+A\leq 1+A+A^2/2!+A^3/3!+\cdots=e^A$$
A: Since
$(e^x)' = e^x$
for all $x$,
and
$e^x > 1$
for
$x > 0$,
we have,
for $x > 0$,
$\begin{array}\\
e^x-1
&=\int_0^x e^t dt\\
&> \int_0^x 1 dt
\qquad\text{since } e^x > 1 \text{ for } x > 0\\
&=x\\
\end{array}
$
A: For $n\in\mathbb{N}$ and $x\gt-1$, we have
$$(1+x)^n\ge1+nx$$
which is easily proved by induction on $n$:
$$\begin{align}
(1+x)^n\ge1+nx\implies(1+x)^{n+1}&=(1+x)(1+x)^n\\
&\ge(1+x)(1+nx)\\
&=1+(n+1)x+nx^2\\
&\ge1+(n+1)x
\end{align}$$
It follows that, if $A\gt0$, then
$$\left(1+{A\over n}\right)^n\ge1+n\left(A\over n\right)=1+A$$
and thus
$$e^A=\lim_{n\to\infty}\left(1+{A\over n}\right)^n\ge\lim_{n\to\infty}(1+A)=1+A$$
Remark:  The inequality actually holds for all $A$, not just $A\gt0$, by the same proof, with the only extra wrinkle being that one needs to note that $A/n\gt-1$ for all $n\gt|A|$.
A: $A+1=e^A$ at $A=0$.
$\frac{d(A+1)}{dA}=1$, while $\frac{d(e^A)}{dA}=e^A$.
$e^A>1 \forall A>0$. 
So they start off equivalent, but the RHS grows faster at all points than the LHS. 
