inductive proof of binomial theorem

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 $n=m+1$,

${(a+b)}^{m+1}$$=$$(a+b){(a+b)}^m$

$=$$(a+b)(a^m+b^m+{\displaystyle \sum _{k=1}^{m-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)a^{m-k}{b}^k)\text{ by the inductive hypothesis}$

$=$$a^{m+1}+b^{m+1}+a{b}^m+b{a}^m+{\displaystyle \sum _{k=1}^{m-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)a^{m-k+1}{b}^k+{\displaystyle \sum _{k=1}^{m-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)a^{m-k}{b}^{k+1}$

$=$$a^{m+1}+b^{m+1}+{\displaystyle \sum _{k=1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)a^{m-k+1}{b}^k+{\displaystyle \sum _{k=0}^{m-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)a^{m-k}{b}^{k+1}\text{ by combining terms}$

$=$$a^{m+1}+b^{m+1}+{\displaystyle \sum _{k=1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)a^{m-k+1}{b}^k+{\displaystyle \sum _{j=1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j-1}}\right)a^{m+1-j}{b}^j\text{ let j=k+1 in second sum}$

$=$$a^{m+1}+b^{m+1}+{\displaystyle \sum _{k=1}^m}\left[\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k-1}}\right)\right]a^{m+1-k}{b}^k\text{ by combining the sums}$

$=$$a^{m+1}+b^{m+1}+{\displaystyle \sum _{k=1}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{m+1}{k}}\right)a^{m+1-k}{b}^k\text{ from Pascal’s rule}$

as desired.