Alternative proofs sought after for a certain identity Here is an identity for which I outlined two different arguments. I'm collecting further alternative proofs, so

QUESTION. can you provide another verification for the problem below?

Problem. Prove that
$$\sum_{k=1}^n\binom{n}k\frac1k=\sum_{k=1}^n\frac{2^k-1}k.$$
Proof 1. (Induction). The case $n=1$ is evident. Assume the identity holds for $n-1$.  Then,
\begin{align*} \sum_{k=1}^{n+1}\binom{n+1}k\frac1k-\sum_{k=1}^n\binom{n}k\frac1k
&=\frac1{n+1}+\sum_{k=1}^n\left[\binom{n+1}k-\binom{n}k\right]\frac1k \\
&=\frac1{n+1}+\sum_{k=1}^n\binom{n}{k-1}\frac1k \\
&=\frac1{n+1}+\frac1{n+1}\sum_{k=1}^n\binom{n+1}k \\
&=\frac1{n+1}\sum_{k=1}^{n+1}\binom{n+1}k=\frac{2^{n+1}-1}{n+1}.
\end{align*}
It follows, by induction assumption, that
$$\sum_{k=1}^{n+1}\binom{n+1}k\frac1k=\sum_{k=1}^n\binom{n}k\frac1k+\frac{2^{n+1}-1}{n+1}=\sum_{k=1}^n\frac{2^k-1}k+\frac{2^{n+1}-1}{n+1}
=\sum_{k=1}^{n+1}\frac{2^k-1}k.$$
The proof is complete.
Proof 2. (Generating functions) Start with $\sum_{k=1}^n\binom{n}kx^{k-1}=\frac{(x+1)^n-1}x$ and integrate both sides: the left-hand side gives
$\sum_{k=1}^n\binom{n}k\frac1k$. For the right-hand side, let $f_n=\int_0^1\frac{(x+1)^n-1}x\,dx$ and denote the generating function
$G(q)=\sum_{n\geq0}f_nq^n$ so that
\begin{align*} G(q)&=\sum_{n\geq0}\int_0^1\frac{(x+1)^n-1}x\,dx\,q^n =\int_0^1\sum_{n\geq0}\frac{(x+1)^nq^n-q^n}x\,dx \\
&=\int_0^1\frac1x\left[\frac1{1-(x+1)q}-\frac1{1-q}\right]dx=\frac{q}{1-q}\int_0^1\frac{dx}{1-(x+1)q} \\
&=\frac{q}{1-q}\left[\frac{\log(1-(1+x)q)}{-q}\right]_0^1=\frac{\log(1-q)-\log(1-2q)}{1-q} \\
&=\frac1{1-q}\left[-\sum_{m=1}^{\infty}\frac1mq^m+\sum_{m=1}^{\infty}\frac{2^m}mq^m\right]=\frac1{1-q}\sum_{m=1}^{\infty}\frac{2^m-1}m\,q^m \\
&=\sum_{n\geq1}\sum_{k=1}^n\frac{2^k-1}k\,q^n.
\end{align*}
Extracting coefficients we get $f_n=\sum_{k=1}^n\frac{2^k-1}k$ and hence the argument is complete.

One can also use the binomial transform.
(If $A(z)=\sum_{i\geq 0} a_i z^i$ is a (formal) power series, the (formal) power series $B(z):=\frac{1}{1-z}
A(\frac{z}{1-z})$ has coefficients $[z^n] B(z)=\sum_{j=0}^n {n \choose j} a_j$).
We have $\log(\frac{1}{1-z})=\sum_{k\geq 1} \frac{z^k}{k}$.
Thus \begin{align*}
\sum_{k=1}^n {n \choose k}\frac{1}{k}&=[z^n] \frac{1}{1-z}\,\log\big(\frac{1}{1-\frac{z}{1-z}}\big)\\
                                     &=[z^n] \frac{1}{1-z}\,\log\big(\frac{1-z}{1-2z}\big)\\
                                     &=[z^n] \frac{1}{1-z}\,\Big(\log\big(\frac{1}{1-2z}\big)-\log\big(\frac{1}{1-z}\big)\Big)\\
                                     &=\sum_{k=1}^n\frac{2^k}{k} -\sum_{k=1}^n \frac{1}{k}\end{align*}
