Proving that the $n$th derivative satisfies $(x^n\ln x)^{(n)} = n!(\ln x+1+\frac12+\cdots+\frac1n)$ Question:
Prove that $(x^n\ln x)^{(n)} = n!(\ln x+1+\frac 12 + ... + \frac 1n)$
What I tried:
Using Leibnitz's theorem, with $f=x^n$ and $g=\ln x$.
So 
$$f^{(j)}=n\cdots(n-j+1)x^{n-j} , g^{(j)}=(-1)^{j+1} \dfrac 1{x^{n-j}}$$
But somehow I get stuck on the way...

Hint: Try using induction. Suppose $(x^n\ln x)^{(n)} = n!\left(\ln x+\frac{1}{1}+\cdots\frac{1}{n}\right)$, then
$$\begin{align}{}
(x^{n+1}\ln x)^{(n+1)} & = \left(\frac{\mathrm{d}}{\mathrm{d}x}\left[x^{n+1} \ln x\right]\right)^{(n)} \\
&= \left((n+1)x^n\ln x + x^n\right)^{(n)} \\
&= (n+1)(x^n\ln x)^{(n)} + (x^n)^{(n)} \\
&= \ldots
\end{align}$$
