Alternative proofs sought after for a certain identity

$\DeclareMathOperator\prob{prob}$Alapan Das' clever argument may be rephrased on the probabilistic language. Write $[m]=\{1,2,\dotsc,m\}$. Choose a random non-empty subset $A\subset [n]$ (all $2^n-1$ possible outcomes having equal probabilities). Then choose a random element $\xi\in A$ uniformly. Denote $p=\prob (\xi=\max(A))$. On one hand, denoting $j=\lvert A\rvert$ we get $$ p=\sum_{j=1}^n \prob(\xi=\max(A)\mathrel||A|=j)\cdot \prob(|A|=j)=\sum_{j=1}^n\frac1j \cdot\frac{{n\choose j}}{2^n-1} $$

On the other hand, $$ p=\sum_{k=1}^n \prob(\xi=k \, \&\,A\subset [k])= \sum_{k=1}^n \prob(\xi=k\mathrel|A\subset [k])\cdot \prob(A\subset [k])\\= \sum_{k=1}^n \frac1k\cdot \frac{2^k-1}{2^n-1}. $$

$$\sum_{k=1}^n\binom nk\frac1k=\sum_{k=1}^n\binom nk\int_0^1 dt\,t^{k-1}= \int_0^1 dt\,\sum_{k=1}^n\binom nk t^{k-1}=\int_0^1 dt\,\frac{(1+t)^n-1}t.$$ $$\sum_{k=1}^n\frac{2^k-1}k=\sum_{k=1}^n\int_0^1 dt\,(1+t)^{k-1}= \int_0^1 dt\,\sum_{k=1}^n (1+t)^{k-1}=\int_0^1 dt\,\frac{(1+t)^n-1}t.$$

\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*}