$$b(a+b)^m = \sum_{j=0}^m \binom{m}{j}a^{m-j}b^{j+1}= \sum_{k=1}^m \binom{m}{k-1}a^{m+1-k}b^{k}+b^{m+1}$$
I believe $j = k-1$ though the book does say that. This is related to proving the binomial theorem inductively. How does the $b^{m+1}$ at the end get there? I would expect that to not be there and $a^{m}b$ to be added before the summation. I must be misunderstanding something fundamental here.
On the one hand: $$\sum_{j=0}^m \binom{m}{j}a^{m-j}b^{j+1} = \binom{m}{0}a^{m}b^{1} + \binom{m}{1}a^{m-1}b^{2} + \dots + \binom{m}{m-1}a^{1}b^{m} + \underbrace{\binom{m}{m}a^{0}b^{m+1}}_{b^{m+1}} $$ On the other hand: $$\sum_{k=1}^m \binom{m}{k-1}a^{m+1-k}b^{k} = \binom{m}{0}a^{m}b^{1} + \binom{m}{1}a^{m-1}b^{2} + \dots + \binom{m}{m-1}a^{1}b^{m}$$
First split off the last term: $$\sum_{j=0}^m\binom mja^{m-j}b^{j+1}=\sum_{j=0}^{m-1}\binom mja^{m-j}b^{j+1}+\binom mma^{m-m}b^{m+1}.$$ Now substitute $k=j+1$ in the first part; the last term simplifies to $b^{m+1}$.
