Let $f(x)=\sum_{k=0}^n \binom{n}{k}\frac{x^{k+1}}{(k+1)^2}$ Note that we have $(xf'(x))'=\sum_{k=0}^n\binom{n}{k}x^k=(1+x)^n$ Therefore we have $xf'(x)=0f'(0)+\int_{0}^x (tf'(t))'dt=\int_{0}^x (1+t)^ndt=\frac{(1+x)^{n+1}-1}{n+1}$ So $f(x)=f(0)+\int_{0}^x \frac{tf'(t)}{t}dt=\int_{0}^x \frac{(1+t)^{n+1}-1}{(n+1)t}dt$. If instead we let $g(x)=\sum_{k=0}^n \frac{x^{k+1}}{k+1}$ we clearly have $g'(x)=\sum_{k=0}^n x^k=\frac{x^{n+1}-1}{x-1}$ and so $g(x)=g(0)+\int_{0}^xg'(t)dt=\int_0^x \frac{t^{n+1}-1}{t-1}dt$. Finally we have $$S=\sum_{k=0}^n \binom{n}{k}\frac{(-1)^{n-k}}{(k+1)^2}+\frac{(-1)^{n+1}}{n+1}g(1)=$$$$(-1)^{n+1}f(-1)+\frac{(-1)^{n+1}}{n+1}g(1)$$$$=(-1)^{1+n}\int_0^{-1}\frac{(1+t)^{n+1}-1}{t}dt+\frac{(-1)^{n+1}}{n+1}\int_0^1 \frac{t^{n+1}-1}{t-1}dt$$$$=\frac{(-1)^n}{n+1}\int_0^1 \frac{u^n-1}{u-1}du+\frac{(-1)^{n+1}}{n+1}\int_0^1 \frac{t^n-1}{t-1}dt=0$$