Find the radius of convergence of the series $y=\sum_{n=0}^{\infty}\binom{p}{n} x^{n}$ Let $p\in R$ Find the radius of convergence of the series: 

$$y=\sum_{n=0}^{\infty}\binom{p}{n} x^{n}$$    

Show that y satisfies the differential equation $(1+x)y'=py$ and initial condition $y(0)=1$



I'm not sure if this question has been asked, but I searched and could not find it. So I know that the radius of convergence is the radius of the largest disk that the series converges in. I'm confused by this problem in general and am unsure of where to even begin. I am looking for a clear solution and explanation to this problem.
It is just the generalization of the usual binomial series:
$$ f(x)=\sum_{n\geq 0}\binom{p}{n} x^n = (x+1)^p.\tag{1} $$
$(1+x)\,f'=p\,f$ is a separable ODE, and $\frac{f'}{f}=\frac{p}{x+1}$ gives that $x=-1$ is a singularity for $f(x)$.

It is the closest singularity to the origin, hence the radius of convergence of $(1)$ is just $\color{red}{1}$.
$$\sum_{n=0}^\infty \binom{p}{n}x^n=\sum_{k=1}^p\binom{p}{k}x^k=(1+x)^p$$
for all $x$. Therefore the radius of convergence is $\mathbb R$.