\begin{align*}
\sum_{k=1}^{n} \frac{2^k-1}{k}
&=\sum_{k=1}^{n} \frac{1}{k}\left(\sum_{j=1}^{k} \binom{k}{j}\right) \\
&=\sum_{j=1}^{n} \sum_{k=j}^{n} \binom{k}{j}\frac{1}{k} \\
&=\sum_{j=1}^{n} \frac{1}{j}\left(\sum_{k=j}^{n} \binom{k-1}{j-1}\right) \\
&=\sum_{j=1}^{n} \frac{1}{j} \binom{n}{j}.
\end{align*}