This can be proved by applying the basic definition and transformation. The left side equals $$\frac{1}{k+1}\binom{n}{k}=\frac{1}{k+1}\cdot\frac{n!}{k!(n-k)!}=\frac{n!}{(k+1)!(n-k)!}$$ The right side equals $$\frac{1}{n+1}\cdot \binom{n+1}{k+1}=\frac{1}{n+1}\cdot\frac{(n+1)!}{(k+1)!(n-k)!}=\frac{n!}{(k+1)!(n-k)!}$$ Hence, the to-be-claimed identity holds.