Solve a function based on an inequality 
Problem: Suppose that $f(x)\in C^\infty (R),~~$and$~~ \forall n\in N,~~x\in R,~~$we have $$|f^{(n)}(x)|\leq e^x$$Prove that $f(x)=f(0)e^x$.

My thought is that prove the derivative of $g=fe^{-x}$ is zero.
However, I can only prove $|g^{(n)}(x)|\leq 2^n$ by mathematical induction and given conditions.
 A: Let $g(x)=e^x-f(x)$, it is obvious that $0\leq g^{(n)}(x)\leq 2e^x$. By Taylor's formula, we have
$$g(x)=g(0)+g'(0)x+\cdots+\frac{g^{(n)}(0)}{n!}x^n+\frac{g^{(n+1)}(\eta_n)}{(n+1)!}x^{n+1},$$
where $\eta_n$ between $0$ and $x$. Since
$$\left|\frac{g^{(n+1)}(\eta_n)}{(n+1)!}x^{n+1}\right|\leq\frac{2e^{\eta_n}}{(n+1)!}|x|^{n+1}\leq\frac{2e^{\max{\{0,x\}}}}{(n+1)!}|x|^{n+1}\to 0$$
as $n\to\infty$, we can get
$$g(x)=\sum_{n=0}^\infty\frac{g^{(n)}(0)}{n!}x^n.$$
The convergence radius of the series is $+\infty$ due to $0\leq g^{(n)}(0)\leq 2$, then $g(z)=\sum_{n=0}^\infty\frac{g^{(n)}(0)}{n!}z^n$ is an entire function. Since (here $a$ is real)
$$|g(z)|=\left|\sum_{n=0}^\infty\frac{g^{(n)}(-a)}{n!}(z+a)^n\right|\leq\sum_{n=0}^\infty\frac{g^{(n)}(-a)}{n!}|z+a|^n=g(|z+a|-a)$$
and 
$$|z+a|-a=\sqrt{(x+a)^2+y^2}-a=\frac{2ax+(x^2+y^2)}{\sqrt{(x+a)^2+y^2}+a}\to x$$
as $a\to+\infty$, we have $|g(z)|\leq g(x)$, where $z=x+iy$. Thus, we obtain
$$|e^{-z}g(z)|\leq e^{-x}g(x)\leq2,$$
that means $e^{-z}g(z)$ is constant by Liouville's theorem, then $e^{-x}f(x)$ is also constant.
