Problem 2-7 in Spivak

Problem 2-7 in Spivak

One is asked to show that $ \sum\limits_{i=1}^{n} k^{p}$ (typo on $i$?)
can always be written in the form
$$\frac{n^{p+1}}{p+1}+An^{p}+Bn^{p-1}+Cn^{p-2}+\cdots.$$
The solution states:
The proof is by complete induction on $p$. The statement is true for
$p=1$, since $$\sum\limits_{k=1}^{n} k= \frac{n(n+1)}{2}=
\frac{n^{2}}{2}+n.$$ Suppose that the statement is true for all natural
numbers $\leq p$. The binomial theorem yield the equations
$$(k+1)^{(p+1)}-k^{p+1}=(p+1)k^{p}+ \textrm{terms involving lower powers
of }k.$$ Adding for $k=1,\dots, n,$ we obtain
$$\frac{(n+1)^{p+1}}{p+1}=\sum\limits_{k=1}^{n} k^{p} + \textrm{terms
involving} \sum\limits_{k=1}^{n} k^{r} \textrm{ for } r<p.$$ By assumption
we can write each $\sum\limits_{k=1}^{n} k^{r}$ as an expression involving
powers $n^{s}$ with $s\leq p$. It follows that $$\sum\limits_{k=1}^{n}
k^{p}=\frac{(n+1)^{p+1}}{p+1} + \textrm{ terms involving powers of }n
\textrm{ less than } p+1.$$
What I tryed based on it and more explicitly:
$$\begin{align}(k+1)^{p+1} &={p+1 \choose p+1}k^{p+1}+{p+1 \choose
p}k^{p}+ \cdots +{p+1 \choose 1}k + {p+1 \choose 0}k^{0}\\ (k+1)^{p+1}&=
1\cdot k^{p+1}+{p+1 \choose p}k^{p}+ \cdots +{p+1 \choose 1}k + {p+1
\choose 0}k^{0}\\ (k+1)^{p+1} -k^{p+1} &= (p+1)k^{p}+ \cdots +(p+1)k +
1\cdot k^{0}\\ \end{align}$$
Adding for $k=1,\dots, n,$ is something like
$$\begin{align} {2^{p+1}} -1^{p+1} &= (p+1)1^{p}+ \cdots +(p+1)1+ 1\cdot
1^{0}\\ {3^{p+1}} - {2^{p+1}} &= (p+1)2^{p}+ \cdots +(p+1)2+ 1\cdot
2^{0}\\ & \vdots\\ (n+1)^{p+1} -{n^{p+1}} &= (p+1)n^{p}+ \cdots +(p+1)n+
1\cdot n^{0}\\ \hline & \hline\\ (n+1)^{p+1} &= (p+1)\sum\limits_{k=1}^{n}
k^{p}+ \cdots +(p+1)\sum\limits_{k=1}^{n} k^{1}+ \sum\limits_{k=1}^{n}
k^{0} + k^{0}\\ \frac{(n+1)^{p+1}}{(p+1)} &= \sum\limits_{k=1}^{n} k^{p}+
\frac{{p+1 \choose p-1}}{(p+1)}\sum\limits_{k=1}^{n}k^{p-1} + \cdots
+\frac{(p+1)}{(p+1)}\sum\limits_{k=1}^{n} k^{1}+
\frac{1}{(p+1)}(\sum\limits_{k=1}^{n} k^{0} + k^{0})\\ \end{align}$$
And assuming the proposition as true for $p-1$ we could write
$$\begin{align} \frac{(n+1)^{p+1}}{(p+1)} &= \sum\limits_{k=1}^{n} k^{p}+
\textrm{terms involving powers of } n\leq p.\\ \sum\limits_{k=1}^{n} k^{p}
&= \frac{(n+1)^{p+1}}{(p+1)}+ \textrm{terms involving powers of } n\leq
p.\\ \end{align}$$
The question is:
How do $(n+1)^{p+1}$ instead of $n^{p+1}$ in the result still makes the
proof valid?