We prove the theorem for a ring. We do not assume a unit for the ring. We do not need commutativity of the ring, but only that $a$ and $b$ commute.

When $n=1$ , the result is clear.

For the inductive step, assume it holds for $m$ . Then for $n = m+1$ , \begin{eqnarray*} (a+b)^{m+1} & = & (a+b)(a+b)^m \\ & = & (a+b)(a^m + b^m+ \sum_{k=1}^{m-1} \binom{m}{k} a^{m-k} b^k )\text{ by the inductive hypothesis} \\ & = & a^{m+1} + b^{m+1} + ab^m + ba^m + \sum_{k=1}^{m-1} \binom{m}{k} a^{m-k+1} b^k + \sum_{k=1}^{m-1} \binom{m}{k} a^{m-k} b^{k+1} \\ & = & a^{m+1} + b^{m+1} + \sum_{k=1}^m \binom{m}{k} a^{m-k+1} b^k + \sum_{k=0}^{m-1} \binom{m}{k} a^{m-k} b^{k+1} \text{ by combining terms} \\ & = & a^{m+1} + b^{m+1} + \sum_{k=1}^m \binom{m}{k} a^{m-k+1} b^k + \sum_{j=1}^m \binom{m}{j-1} a^{m+1-j} b^j \text{ let j=k+1 in second sum} \\ & = & a^{m+1} + b^{m+1} + \sum_{k=1}^m \left[ \binom{m}{k} + \binom{m}{k-1} \right] a^{m+1-k}b^k \text{ by combining the sums}\\ & = & a^{m+1} + b^{m+1} + \sum_{k=1}^m \binom{m+1}{k} a^{m+1-k}b^k \text{ from Pascal's rule} \\ \end{eqnarray*}as desired.