Here's a probabilistic approach copied from my answer to a duplicate question. I am reposting here (marked as CW) for the record. We want to show that $ \frac{1}{2^{n+1}-1} \sum_{j=0}^n \binom{n}{j} \frac{1}{j+1} = \frac{1}{n+1}. \tag{$\ast$} $
Consider an experiment where we pick a nonempty subset ("committee") $S \subseteq \{ 0, 1, 2, \ldots, n \}$ uniformly at random and pick an $x$ uniformly at random from $S$ (the "head" of the committee). Then both sides count the probability that the head is $0$. By symmetry the head is a uniformly random person, so it is $n+1$ with probability $\frac{1}{n+1}$. Therefore, it only remains to justify the left hand side of $(\ast)$.
The probability of the event $E_j$ that $S$ contains $0$ and it also $j$ other people from $\{ 1, 2, \ldots, n\}$ is $ \frac{1}{2^{n+1}-1} \binom{n}{j}. $ Conditioned on $E_j$, the probability that the head is $0$ is equal to $\frac{1}{j+1}$. [Finally, conditioned on the event that $0 \notin S$, the probability that the head is $0$ is zero.] Thus, by the law of total probability, $ \begin{align*} \Pr[\text{Head is } 0] &= \sum_{j=0}^n \Pr[E_j] \cdot \Pr[\text{Head is } 0 \mid E_j] \\ &= \sum_{j=0}^n \frac{1}{2^{n+1}-1} \binom{n}{j} \cdot \frac{1}{j+1} \end{align*} $