From # 3157, we know that $$\sum_{k=1}^{n}(-1)^k\binom{n}{k}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$$

Let $$a_0 = 0, a_n = 1+\frac{1}{2}+\cdots+\frac{1}{n} \quad, n=1, 2, \cdots$$

and $$b_0=0, b_n = -\frac{1}{n}\quad, n=1, 2, \cdots$$

Then we have already known that $$a_n = \sum_{k=0}^{n}(-1)^k\binom{n}{k}b_n$$

Then utilizing the inverse method immediately gives us $$b_n = \sum_{k=0}^{n}(-1)^k\binom{n}{k}a_n$$

which is the to-be-proved claim.