Why $e^x$ never equal $x$? Je veux savoir pourquoi $x=e^x$ n'a aucune solution dans $\Bbb R$. Lorsque j'ai essayé de tracer le graphe de la fonction $e^x$, j'ai trouvé en fait qu'elle est une fonction strictement croissante mais je ne sais pas quoi faire après. 
Merci mon ami.

I would like to know why $x=e^x$ has no solution in $\Bbb R$? I tried to plot the function $e^x$ and I found that it is an increasing function but I do not know what to do next.
 A: If we look at the initial value for each function $f(x)=x$ and $g(x)=e^x$ we see that $f(0)=0 < g(0)=1$.
Now for $x>0$ we have $f'(x)=1$ and $g'(x) = e^x$ which tells us that $f'(x) < g'(x)$ for all $x > 0$, and so $$x \le e^x$$ for all $x \ge 0$. For negative $x$ we know that $e^x$ is strictly positive, so this tells us that $$x \le e^x$$ for all $x \in \mathbb{R}$.
A: The exponential function is a convex function, since the second derivative equals the function, so it is non-negative. This implies that the graphics of $f(x)=\exp(x)$ lies above the tangent line in $x=0$, i.e.:
$$\forall \in\mathbb{R},\quad e^x \geq x+1.$$
A: This is easily checked, by definition of $e^x$. For $x > 0$ we have 
$$ e^x = 1 + x + \sum_{k = 2}^\infty\frac{x^k}{k!}>x$$ and for $x \leq 0$ we have $$x \leq 0 < e^x.$$
A: If $x \leq 0$, then $e^x > 0 \geq x$. 
Suppose $x > 0$ and let $f(x) = e^x - x $. So $f'(x) = e^x - 1 > e^0 - 1 = 0$. Hence $f$ is increasing and since $f(0) > 0$, $f(x) = e^x - x > 0$, finally $e^x > x$.
The inequality is strict.
A: By Bernoulli's inequality,
$$ \left(1+\frac xn\right)^n \ge 1+x \qquad\text{if $n > |x|$.} $$
Letting $n\to\infty$ yields $e^x \ge 1+x > x$.
(I like this argument because it doesn't need to treat $x<0$ specially.)
