Algebraic expansion of $1 + (1 + x) + ... + (1 + x)^{p-1}$ I was proving that $p(x) = 1 + x + \text{ ... }  + x^{p-1}$, where $p$ is a prime number, is irreducible over the rationals, by using the translation $1 + (1 + x) + \text{ ... } + (1 + x)^{p-1}$. I managed to do it, but I would like to get a closed expression for this as a polynomial.
So far, I have:
\begin{equation}
\sum_{k = 0}^{p-1}{(1+x)^k} = \sum_{k = 0}^{p-1}{(\sum_{i = 0}^k{\binom{k}{i}x^i)}} 
\end{equation}
By inspection, it looks like the final result will be:
\begin{equation}
\sum_{k = 0}^{p-1}{(1+x)^k} = \sum_{k = 0}^{p-1}{(\sum_{i=k}^{p-1}{\binom{i}{k}}} )x^k 
\end{equation}
I.E.
\begin{equation}
\sum_{i=0}^{p-1}{\binom{i}{0}} + \sum_{i=1}^{p-1}{\binom{i}{1}} x + \text{ ... } + \sum_{i=p-1}^{p-1}{\binom{i}{p-1}}x^{p-1} 
\end{equation}
First of all, is this correct? If not, what would be the correct expression and why? And if it is correct, how can I get from the first equation to the second?
Thanks in advance!
 A: Assuming that I have not misinterpreted your query:
A case can be made that the following hint-answer is defective, because it avoids examining your work.  Through no fault of your own, you went down a natural but bad path.
I actually don't know how to prove that any polynomial is irreducible, but I can give you a hint about how to very easily find a closed form expression for the requested polynomial.
Hint-1:
Given the geometric series, 
$f(t) = 1 + t + t^2 + \cdots + t^n$, what is the result of 
$f(t) \times (1-t)$?
Hint-2:
What happens if you set $t = (x+1)$?
A: If $p\in\Bbb P$ then for $0\le k\le p-1$ the $x^k$ coefficient in $\sum_{j=0}^{p-1}(1+x)^j$ is $\sum_{j=k}^{p-1}\binom{j}{k}=\binom{p}{k+1}$ by the hockey-stick identity. This is presumably how you proved irreducibility by Eisenstein's criterion, since the non-leading coefficients will be multiples of $p$. This implies the closed form$$\sum_{k=0}^{p-1}\binom{p}{k+1}x^k=\frac1x\sum_{\ell=1}^p\binom{p}{\ell}x^\ell=\frac{(1+x)^p-1}{x},$$but that's more easily obtained by @user2661923's strategy. Both approaches make obvious that this closed form is true even for composite $p$.
