This question is motivated by the MO problem here. Perhaps it is not that difficult.

Question.Here is an cute formula. $$\frac1n\sum_{k=0}^{n-1}\frac1{\binom{n-1}k}=\sum_{k=1}^n\frac1{k2^{n-k}}.$$ I've one justification along the lines of Wilf-Zeilberger (see below). Can you provide an alternative proof? Or, any reference?

The claim amounts to $a_n=b_n$ where $$a_n:=\sum_{k=0}^{n-1}\frac{2^n}{n\binom{n-1}k} \qquad \text{and} \qquad b_n:=\sum_{k=1}^n\frac{2^k}k.$$ Define $F(n,k):=\frac{2^n}{n\binom{n-1}k}$ and $\,G(n,k)=-\frac{2^n}{(n+1)\binom{n}k}$. Then, it is routinely checked that $$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k),\tag1$$ for instance by dividing through with $F(n,k)$ and simplifying. Summing (1) over $0\leq k\leq n-1$: \begin{align} \sum_{k=0}^{n-1}F(n+1,k)-\sum_{k=0}^{n-1}F(n,k) &=a_{n+1}-\frac{2^{n+1}}{n+1}-a_n, \\ \sum_{k=0}^{n-1}G(n,k+1)-\sum_{k=0}^{n-1}G(n,k) &=-\sum_{k=1}^n\frac{2^n}{(n+1)\binom{n}k}+\sum_{k=0}^{n-1}\frac{2^n}{(n+1)\binom{n}k}=0. \end{align} Therefore, $a_{n+1}-a_n=\frac{2^{n+1}}{n+1}$. But, it is evident that $b_{n+1}-b_n=\frac{2^{n+1}}{n+1}$. Since $a_1=b_1$, it follows $a_n=b_n$ for all $n\in\mathbb{N}$.

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Here is an alternative approach to Fedor's answer below in his elaboration of Fry's comment.

With $\frac1{n+1}\binom{n}k=\int_0^1x^{n-k}(1-x)^kdx$, we get $a_{n+1}=2^{n+1}\int_0^1\sum_{k=0}^nx^{n-k}(1-x)^kdx$. So, \begin{align} 2^{n+1}\int_0^1 dx\sum_{k=0}^nx^{n-k}(1-x)^k &=2^{n+1}\int_0^1x^ndx\sum_{k=0}^n\left(\frac{1-x}x\right)^k \\ &=2^{n+1}\int_0^1x^n\frac{\left(\frac{1-x}x\right)^{n+1}-1}{\frac{1-x}x-1}dx \\ &=\int_0^1\frac{(2-2x)^{n+1}-(2x)^{n+1}}{1-2x}\,dx:=c_{n+1}. \end{align} Let's take successive difference of the newly-minted sequence $c_{n+1}$: \begin{align} c_{n+1}-c_n &=\int_0^1\frac{(2-2x)^{n+1}-(2-2x)^n+(2x)^n-(2x)^{n+1}}{1-2x}\,dx \\ &=\int_0^1\left[(2-2x)^n+(2x)^n\right]dx=2^{n+1}\int_0^1x^ndx=\frac{2^{n+1}}{n+1}. \end{align} But, $b_{n+1}-b_n=\frac{2^{n+1}}{n+1}$ and hence $a_n=b_n$.