Proceed by weak induction on $n$ over $\mathbb{N}$. The case $n = 0$ follows from the result R5671: Extreme values for binomial coefficient since \begin{equation} (x + y)^0 = 1 = \binom{0}{0} = \sum_{m = 0}^0 \binom{0}{m} x^m y^{-m} \end{equation} Suppose then that the claim holds for some $n \geq 0$. Using result R5670: , we have \begin{equation} \begin{split} (x + y)^{n + 1} & = (x + y) (x + y)^n \\ & = (x + y) \sum_{m = 0}^n \binom{n}{m} x^m y^{n - m} \\ & = x \sum_{m = 0}^n \binom{n}{m} x^m y^{n - m} + y \sum_{m = 0}^n \binom{n}{m} x^m y^{n - m} \\ & = \sum_{m = 0}^n \binom{n}{m} x^{m + 1} y^{n - m} + \sum_{m = 0}^n \binom{n}{m} x^m y^{n + 1 - m} \\ & = \sum_{m = 1}^n \binom{n}{m - 1} x^{m} y^{n + 1 - m} + \sum_{m = 1}^n \binom{n}{m} x^m y^{n + 1 - m} + \binom{n}{0} x^0 y^{n + 1} \\ & = \sum_{m = 1}^n \binom{n + 1}{m} x^{m} y^{n + 1 - m} + \binom{n}{0} x^0 y^{n + 1} \\ & = \sum_{m = 1}^n \binom{n + 1}{m} x^{m} y^{n + 1 - m} + \binom{n + 1}{0} x^0 y^{n + 1} \\ & = \sum_{m = 0}^n \binom{n + 1}{m} x^{m} y^{n + 1 - m} \\ & = \sum_{m = 0}^{n + 1} \binom{n}{m} x^{m} y^{n - m} \\ \end{split} \end{equation} The claim now follows from R5104: Proof by principle of weak mathematical induction on the natural numbers. $\square$