Why does $(1+x)^np(x)$ has at least $n+1$ terms? $p(x)$ is a polynomial. Assume $p(x)\ne0$. 
How to prove that $(1+x)^np(x)$ has at least $n+1$ terms?
 A: Hint: Induction with differentiation to get to smaller step.
A: For a polynomial $f\in\mathbb C[X]$, $f(x)=\sum_ka_kX^k$ let $$w(f)=|\{\,k\in\Bbb N:a_k\ne 0\,\}|$$ denote the number of nonzero terms occurring in $f$ and let $$v(f)=\begin{cases}\max\{\,k\in\Bbb N:(1+X)^k\mid f\,\}&\text{if $f\ne 0$}\\\infty&\text{if $f=0$}\end{cases}$$
denote the multiplicity of $-1$ as a root of $f$.
Observe that $$\tag1w(Xf)=w(f)$$ and that $$\tag2w(f')=\begin{cases} w(f)&\text{if $X\mid f$}\\w(f)-1&\text{otherwise.}\end{cases}$$
Also,
$$\tag3v(Xf)=v(f) $$
and as the derivative of $(1+X)^kq(X)$ with $q(-1)\ne 0$ is $(1+X)^{k-1}r(X)$ with $r(X):=kq(X)+(1+X)q'(X)$ and $r(-1)=kq(-1)$ we have
$$\tag4v(f')=v(f)-1\qquad\text{if $0<v(f)<\infty$} $$
The problem statement is equivalent to saying that $w(f)>v(f)$ for all nonzero $f$.
This follows readily by induction, the base step being constant polynomials where $w(f)=1>0=v(f)$, and the induction step using $(1)$ and $(3)$ if $X\mid f$ and  $(2)$ and $(4)$ if $X\nmid f$.
A: If p(x) is non-zero then it  contains at least one term in it. How many terms are in $(1+x)^n$? The binomial expansion will help here.
