Show that not exists any polynomial function such that $f(x) = \log (1+x)$. Does anyone have any idea on that problem?


  Let $f : \mathbb{R} \to \mathbb{R}$ be a polynomial function. Show that not exists any $f$ such that $f(x) = \log (1+x)$.


It's easy to show that $a_0 = 0$ and $a_1 = 1$. But after i don't have any idea. Any point? Thanks!
Maybe there are many ways to show this; but what I have in mind is this:
If $f$ is a polynomial of degree $k$, then its $(k+1)$-th derivative is zero. 
But this is not the case for $\log(1+x)$; the $(k+1)$-th derivative of $\log(1+x)$ does not vanish.
If $\log(1+x)$ is polynomial, $f$ is a transcendental polynom function which doesn't exist.