Proof of $a^x ≥ x+1 \; \forall x \in \Bbb R \implies a=e$ I'm trying to prove the following :

Let $a>0$ a real number. Then : 
  $\quad a^x ≥ x+1 \;\; \forall x \in \Bbb R \iff a=e$

I managed to prove the '$\Longleftarrow$' part : $x≥0$ then $e^x≥x+1$ is clear, knowing that $e^x = \sum\limits_{n≥0} x^n/n!$. On the other hand, if $x<0$, then $1-e^x = \int_x^0 e^tdt ≤ \int_x^0 1 = -x$.

Here is what I've done for the '$\Longrightarrow$' part. Under the given hypothesis, it is easy to show that $a>1$. Let $f_a(x)=a^x-(x+1) ≥ 0$.
Then $$f_a'(x)=\ln(a)a^x-1=0 \iff x=-\log_a(\ln(a))$$
It is easy to show that $m(a) = -\log_a(\ln(a))$ is a minimum for $f_a$. In particular,
$$\begin{align}
f_a(m(a))≥0 \iff \\
\frac{1}{\ln(a)} - (m(a)+1)≥0 \iff \\
\frac{1}{\ln(a)} ≥ 1- \frac{\ln(\ln(a))}{\ln(a)} \underset{a>1}\iff \\
1≥\ln(a)-\ln(\ln(a)) \iff \\
e≥\frac{a}{\ln(a)} \iff \\
h(a)=e\ln(a)-a≥0
\end{align}$$
Since $h'(a)=\frac{e}{a}-1=0 \iff a=e$ and $h''(a)<0$, it is easy to see that $h$ has a unique maximum at $a=e$, and then $h(a)≤h(e)=0, h(a)=0 \iff a=e$. Therefore, I conclueded that $a=e$ as desired.

Is it possible to do an easier proof, or is my proof fine ?

 A: Without using derivation, one can use the inequality $\ln(x)\leq x-1$ and its equivalent form that:
$$
\ln(x)\geq 1-\frac{1}x. (\star)
$$
Now suppose that there is $m$ such that $\forall x$ $\ln(x)\leq m(x-1)$. Note that this is equivalent to your problem. Indeed $m=\ln(a)$ and we can safely assume that $m>0$ (put $x=1$ and then $a>2$). So we have:
$$
(1) \ln(x)\leq m(x-1)
$$
and using $(\star)$ we have:
$$
1-\frac 1x\leq mx-m \implies mx+\frac 1x\geq m+1.
$$
But choosing $x=\frac1{\sqrt m}$ shows that the previous inequality turns into:
$$
2\sqrt m\geq m+1\implies (\sqrt m -1)^2\leq 0.
$$
So $m$ should be one.
A: If you set $t=x+1$, the inequality becomes $a^{t-1}\ge t$ and it's sufficient to analyze it for $t>0$, so it is the same as
$$
(t-1)\log a\ge\log t
$$
Consider the function
$$
f(t)=(t-1)\log a-\log t
$$
We want to see whether or not the function is nonnegative. If $0<a\le1$ it's obvious it isn't, since in this case
$$
\lim_{t\to\infty}f(t)=-\infty
$$
So we assume $a>1$. We have
$$
\lim_{x\to0}f(t)=\infty=\lim_{t\to\infty}f(t)
$$
and
$$
f'(t)=\log a-\frac{1}{t}=\frac{t\log a-1}{t}
$$
which shows $f$ has a minimum at $t=1/\log a$. Since
$$
f(1/\log a)=1-\log a+\log\log a=\log\frac{e\log a}{a}
$$
the function is nonnegative if and only if
$$
e\log a\ge a
$$
The function $g(u)=e\log u-u$ has derivative
$$
g'(u)=\frac{e-u}{u}
$$
which vanishes at $u=e$, which is an absolute maximum, and $g(e)=0$. So $e\log a\ge a$ holds if and only if $a=e$.
