To finish off Yuval's argument, we try an inductive proof on
$\sum_{k=1}^n \binom{n}{k}\frac{(-1)^k}{k}=-\sum_{k=1}^n \frac1{k}=-H_n$
The equality is easily verified for $n=1$. We now try to simplify
$\sum_{k=1}^{n+1} \binom{n+1}{k}\frac{(-1)^k}{k}-\sum_{k=1}^n \binom{n}{k}\frac{(-1)^k}{k}$
which goes a bit like this:
$\begin{align*} &\frac{(-1)^{n+1}}{n+1}+\sum_{k=1}^n \binom{n+1}{k}\frac{(-1)^k}{k}-\sum_{k=1}^n \binom{n}{k}\frac{(-1)^k}{k}\\ &\frac{(-1)^{n+1}}{n+1}+\sum_{k=1}^n \left(\binom{n+1}{k}-\binom{n}{k}\right)\frac{(-1)^k}{k}\\ &\frac{(-1)^{n+1}}{n+1}+\frac1{n+1}\sum_{k=1}^n (-1)^k\binom{n+1}{k}\\ &\frac{(-1)^{n+1}}{n+1}+\frac1{n+1}\left(-1-(-1)^{n+1}+\sum_{k=0}^{n+1} (1)^{n-k+1}(-1)^k\binom{n+1}{k}\right)\\ &\frac{(-1)^{n+1}}{n+1}+\frac1{n+1}\left(-1-(-1)^{n+1}+(1-1)^{n+1}\right)=-\frac1{n+1}\\ \end{align*}$
and since we also have $-H_{n+1}-(-H_n)=-\dfrac1{n+1}$, the equality is established.
Here's an explicit application of Leibniz's product rule for the second problem, which uses the identity $\dfrac{\mathrm d^n}{\mathrm dx^n}\dfrac1{x}=\dfrac{(-1)^n n!}{x^{n+1}}$:
$\begin{align*} \frac{\mathrm d^n}{\mathrm dx^n}\frac{\ln\,x}{x}&=\frac{(-1)^n n!}{x^{n+1}}\ln\,x+\sum_{k=1}^n\binom{n}{k}\left(\frac{\mathrm d^k}{\mathrm dx^k}\ln\,x\right)\left(\frac{\mathrm d^{n-k}}{\mathrm dx^{n-k}}\frac1{x}\right)\\ &=\frac{(-1)^n n!}{x^{n+1}}\ln\,x+\sum_{k=1}^n\frac{n!}{\color{blue}{k!}\color{magenta}{(n-k)!}}\left(\frac{\color{red}{(-1)^{k-1}}\color{blue}{(k-1)!}}{\color{orange}{x^k}}\right)\left(\frac{\color{red}{(-1)^{n-k}}\color{magenta}{(n-k)!}}{\color{orange}{x^{n-k+1}}}\right)\\ &=\frac{(-1)^n n!}{x^{n+1}}\ln\,x+\frac{\color{red}{(-1)^{n-1}} n!}{\color{orange}{x^{n+1}}}\sum_{k=1}^n\frac1{\color{blue} k}=\frac{(-1)^n n!}{x^{n+1}}\left(\ln\,x-\sum_{k=1}^n\frac1{k}\right) \end{align*}$