Answer
$p$ is a nonzero integer
Work Step by Step
We are given the the function $f(x)=(1+x)^p$. The Taylor series centered at 0 is:
$f(x)=f(0)+f'(0)x+\dfrac{f"(0)}{2!}x^2+........$
We have:
$f'(x)=p(1+x)^{p-1}\Rightarrow f'(0)=p$
$f"(x)=p(p-1)(1+x)^{p-2}\Rightarrow f"(0)=p(p-1)$
$f^{(3)}(x)=p(p-1)(p-2)(1+x)^{p-3}\Rightarrow f^{(3)}(0)=p(p-1)(p-2)$
......
The derivative of order $p$ is:
$f^{(p)}(x)=p(p-1)(p-2)....\cdot 2\cdot 1(1+x)^{p-p}=p!\Rightarrow f^{(p)}(0)=p(p-1)(p-2)....\cdot 2\cdot 1=p!$
The derivatives of order greater than $p$ are all zero, therefore teh series terminate.
The condition for the series to terminate is that the exponent $p$ is a nonzero integer.
