Problem: Show that $(1+x)^p$ converges everywhere for $p \in \mathbb{N}$, and for $|x| < 1$ otherwise.
My work: I think that if $p \in \mathbb{N}$ then the Taylor series will just be a polynomial equal to the function so will converge everywhere.
To work out the radius of convergence, consider the $n$-th and $n+1$-th terms, given by $A_n$ and $A_{n+1}$.
$A_n = \frac {p(p-1) \cdots (p-n+1)}{n!} x^n$
$A_{n+1} = \frac {p(p-1) \cdots (p-n+2)}{(n-1)!} x^{n-1}$
So $\frac {|A_{n+1}|}{|A_n|} = \frac {|x|} {n(p-n+2)}$
$\lim_{n \rightarrow \infty} = 0$ therefore the Taylor series will converge everywhere.
How does this proof work for $p \not \in \mathbb{N}$?
You are presumably asking about the convergence of the Maclaurin series for $(1+x)^p$.
Your argument for non-negative integers $p$ is correct. The series is a polynomial. That's all that needs to be said. Suppose now that $p$ is not a non-negative integer.
The Ratio Test was not applied in quite the right way. The coefficient $a_n$ of $x^n$ is given by
$$a_n=\frac{p(p-1)\cdots(p-n+1)}{n!},$$ and therefore $$a_{n+1}=\frac{p(p-1)\cdots(p-n)}{(n+1)!}.$$ Thus the absolute value of $\frac{a_{n+1}}{a_n}$ is $$\frac{|p-n|}{n+1}.$$ This has limit $1$ as $n\to\infty$. So we have divergence if $|x|\gt 1$ and convergence if $|x|\lt 1$.
Remark: Note that we have not dealt with $x=\pm 1$. It is not clear from the wording of the problem whether we are expected to. But in fact if $p$ is not a non-negative integer, then the Maclaurin series of $(1+x)^p$ does not converge at $x=\pm 1$. One can show this by showing that the terms do not approach $0$.