$(n + 1)^k$ number of different degrees of $n$ How can I get the number of different degrees of $n$
that would generate from $(n + 1)^k$.
For example, for $k = 2$,
I want to get $3$ -- ($n^2+2n +1$).
Thanks
 A: the proof:
For $k=2$ we have the number of different degrees of $n$ is $3$ (because $(n+1)^2=n^2+2n+1$), we suppose that the number of different degrees of $n$ is $k+1$ for $(n+1)^k$,
so we have:
$$(n+1)^{k+1}=(n+1)^{k+1}(n+1)=n(n+1)^{k+1}+(n+1)^{k+1}$$
by hypothesis we know that the number of different degrees of $n$ of $(n+1)^k$ is $k+1$,  multiplying by $1$ we get the same elements of the same degrees and by multiplying by $n$ we get another different degree which does not exists is the degree $k+1$ (of $n^{k+1}$), so we conclude that the number of different degrees of $(n+1)^{k+1}$ is $k+2$.
Finally we conclude by induction principle that the number of different degrees of $(n+1)^k$ is $k+1$.
A: $$(n+1)^k=\underset{k\text{ factors of }(n+1)}{\underbrace{(n+1)(n+1)\cdots(n+1)}}$$
In expanding that product, the terms will come from choosing either $n$ or $1$ from each of the $k$ factors.  If we choose all $1$s, we'll get $n^0$; if we choose all $n$s, we'll get $n^k$;  if we choose $p$ $n$s and $k-p$ $1$s (for $0\le p\le k$), we'll get $n^p$.  So there are terms with $n^p$ for $p=0,1,\dots,k$, or $k+1$ distinct powers of $n$.
