$f(x) = \ln(x)f'(x)$. I would like to find a differentiable $f:[1,\infty) \to \mathbb{R}$ satisfying: 
$$f(x) = \ln(x)f'(x)$$.
My work:

We have $f(1) = 0$ and $f(e)=f^k(e)$. A Taylor series centered at $x_0=e$ gives us the following polynomial:
  $$f(e)+f(e)(x-e)+f(e)\frac{1}{2}(x-e)^2 \dots$$
$$=f(e) \sum_{k=1}^\infty \frac{(x-e)^k}{k!}$$
  although I don't see how this might help me, and I also am not sure how else to approach this problem. Does the condition $f(e)=f^k(e)$ force $f(x)$ to be $e^x$, which would contradict $f(x)=\ln(x)f'(x)$, and thus no $f$ exists? I wouldn't think so, although I know that $Ae^x$ is the unique solution to $f(x)=f'(x)$ from this post.

 A: As Caffeine commented
$$\frac{f'(x)}{f(x)}=\frac1{\log(x)}\implies f(x)=C e^{\text{li}(x)}=C e^{\text{Ei}(\log(x))}$$
If you look for a series expansion of $e^{\text{li}(x)}$ around $x=e$, it is given by
$$e^{\text{Ei}(1)}\left(1+(x-e)+\frac{(e-1) }{2 e}(x-e)^2+\frac{e^2-3e+3}{6 e^3}(x-e)^3+ O\left((x-e)^4\right)\right)$$ where $e^{\text{Ei}(1)}\approx 6.65333$
A: As written, the equation has a nasty singularity at $x=1$ and this is reflected in the solution with the integral logarithm (see Claude's answer).
Around $x=1$ we can write $\log x\approx x-1$, leading to the solution $f(x)\approx|x-1|$. Hence we substitute $f(x)=(x-1)g(x)$, giving us
$$\frac{g'(x)}{g(x)}=1-\frac{\log x}{x-1}.$$
With the kind help of WA,
$$\log g(x)+c=x+\text{Li}_2(1-x)=x+\sum_{k=1}^\infty\frac{(1-x)^k}{k^2}.$$
($\text{Li}_2$ is the dilogarithm function, which is quite smooth.) The given series converges in $[0,2]$. Values elsewhere can be obtained by these identities: https://en.wikipedia.org/wiki/Spence%27s_function#Identities.
