trouble with proof of the binomial theorem Below is Spivak's solution to his question to prove the binomial theorem. It is  solution (d). From 'Then', there are 4 steps. I could only understand the first two. I do not understand how j is replaced by j-1 and how n became n+1. It would really help if someone were to suggest intermediate steps to arrive at that result.

The binomial theorem is clear for $n=1$. Suppose that
$$(a+b)^n=\sum_{j=0}^n{n\choose j}a^{n-j}b^j$$
Then
\begin{align}
(a+b)^{n+1}&=(a+b)(a+b)^n=(a+b)\sum_{j=0}^n{n\choose j}a^{n-j}b^j\\
&=\sum_{j=0}^n{n\choose j}a^{n+1-j}b^j+\sum_{j=0}^n{n\choose j}a^{n-j}b^{j+1}\\
&=\sum_{j=0}^n{n\choose j}a^{n+1-j}b^j+\sum_{j=1}^{n+1}{n\choose j-1}a^{n+1-j}b^j\ \text{(replacing }j \text{ by } j-1 \text)\\
&=\sum_{j=0}^{n+1}{n+1\choose j}a^{n+1-j}b^j
\end{align}

 A: It might be helpful to write out the first few terms of the sums $\displaystyle\sum_{j=0}^n {n\choose j} a^{n-j}b^{j+1}$ and $\displaystyle\sum_{j=1}^{n+1} {n\choose j-1} a^{n+1-j}b^{j}$ to convince yourself that they are equivalent. 
Alternatively, note that in both expressions $j$ is a dummy variable that has no significance outside the sigma notation. It might be helpful, then, to give the dummy variable it a different name, say $u$,  in the second sum.
That is, for the sum
$$\displaystyle\sum_{j=0}^n {n\choose j} a^{n-j}b^{j+1}$$ 
Let $u=j+1$, then when $j=0$, $u=1$; when $j=n$, $u=n+1$. Then just express the $j$ in the general term appropriately in terms of $u$. You should get:
$$\displaystyle\sum_{u=1}^{n+1} {n\choose u-1} a^{n+1-u}b^{u}$$
But since $u$ is a dummy variable, we could as easily have called it $j$.
A: Let $j=i-1$ so that when $j$ goes from $0\dots n\,$, $i=j+1$ goes from $1 \dots n+1$. Then:
$$
\sum_{j=0}^{n} \binom{n}{j} a^{n-j}b^{j+1} = \sum_{i=1}^{n+1} \binom{n}{i-1} a^{n-(i-1)}b^{(i-1)+1} =  \sum_{i=1}^{n+1} \binom{n}{i-1} a^{n+1-i}b^{i}
$$
Of course, you can replace $i$ with $j$ in the last sum, since the index of the sum is just a placeholder.
