Reduction formulas for higher order derivations and a hypergeometric identity

A derivation (of order 1) satisfies the reduction formula $$f(x^k) = kx^{k-1}f(x)$$ for any integer k. In this article we find corresponding reduction formulas for derivations of higher order on commutative rings. More precisely, for every derivation f of order n and every positive integer k we find an explicit formula for $$f(x^k)$$ as a linear combination of $$x^{k-1}f(x),x^{k-2}f(x^2), \ldots , x^{k-n}f(x^n)$$ . The proof hinges on the hypergeometric identity $$\begin{aligned} \sum _{k \ge 0} (-1)^k \left( {\begin{array}{c}n\\ k\end{array}}\right) \left( {\begin{array}{c}n+d+1-k\\ n-j\end{array}}\right) \left( {\begin{array}{c}d+j-k\\ j\end{array}}\right) = \left( {\begin{array}{c}n\\ j\end{array}}\right) \end{aligned}$$ for any positive integer n, nonnegative integer d, and integer j satisfying $$0 \le j \le n$$ . We prove this identity via the WZ-method.