For arbitrary $k$, the numbers in set $A_j = \{ f(k(i+1)+j)-f(ki+j) \ | \ 0 \leqslant i \leqslant n-2 , \ 1 \leqslant j \leqslant k \}$ are multiple of $k$. So we have as choices as $k$ multiples in range. Suppose we have $x$ elements in range that are multiples of $k$. So we have $x^{|A_j|} = x^{n-1}$ choices for $A_j$. Since we have $k$ number of $A_j$ that are disjoint any two of them. So we have $(x^{n-1})^k$. For last step, suppose we have $y$ elements in range at all. Now for every $A_1, \dots, A_k$ there is $y$ different choices. So we have: $$(yx^{n-1})^k$$different choices for function $f$, that is a perfect $k$-power. This answer is a generalization for arbitrary $k$ and range set.