Suppose $f$ has derivatives of all orders. Prove that $F(x):=exp(f(x))$ also has derivatives of all orders. Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ has derivatives of all orders. 
Prove that $F(x):=exp(f(x))$ also has derivatives of all orders.

Genuinely very confused by this question. I used an induction, but it looks really sloppy and I want some criticism. First, define the $nth$ derivative by $f^{(n)}$
I note that $exp^1(f(x))=exp(f(x))$. Thus, for $n=1$, $exp(f(x))$ is differentiable. To prove infinite differentiability, we need to show that $exp(f(x))$ is differentiable for all $n$. 
Suppose $exp^n(f(x))=exp(f(x))$ and is differentiable. Then, consider $exp^{(n+1)}(f(x))$.
$exp^{(n+1)}(f(x))=exp^n(f(x))=exp(f(x))$
Thus, $exp(f(x))$ is differentiable for all $n$, i.e. is infinitely differentiable.

This seems so naive. I know that its wrong. Please help me with a different method.
 A: A slick way to do this by induction is as follows: suppose the $k$th derivative exists for all $0 \le k \le n$.  Then
$$ \frac{d^{n+1}}{dx^{n+1}} e^{f(x)} = \frac{d^{n}}{dx^{n}} (f'(x) e^{f(x)}) = \sum_{k=0}^n \binom nk f^{(k+1)}(x) \frac{d^{n-k}}{dx^{n-k}} e^{f(x)} ,$$
where the last step follows by Leibnitz' formula.
I saw this trick in this paper:
Ma, W.T., Sandri, G.Vh. and Sarkar S., (1992) Analysis of the Luria-Delbrück Distribution Using Discrete Convolution Powers. Journal of Applied Probability, 29, 255-267.
A: Look again at your statements regarding the calculation of derivatives of F.  
Per walkar you'll want to convince yourself that F'(x) = F(x)f'(x). Next, using this fact, if you work out the first several derivatives of F, writing each one as a sum of products of derivatives of F and f, you'll notice a pattern in the coefficients and the orders of the derivatives in each term of the sum.  The patterns will suggest a binomial-like expansion is possible for each derivative i.e. F', F'', F''', etc. can each be written as a binomial expansion of terms involving an ith derivative of F and a jth derivative of f. 
Once you've established this you can use induction to state that if the first n-1 derivatives of F exist, then the nth derivative of F can be written in terms of the previous n-1, proving that the nth derivative exists as well.  All the derivatives of f are stated to exist so its really the derivatives of F we care about.
