Certainly. $\displaystyle \frac{x^a-1}{x-1}$ has the form $\displaystyle x^{a-1} + x^{a-2} + \cdots + x + 1$, correct? So what you're really asking for is the set of integers of the form

$\displaystyle \alpha = \sum_{i=0}^{n} x^n$

which are such that $\displaystyle \forall$ primes $\displaystyle p$, $\displaystyle p|\alpha$ $\displaystyle \Rightarrow$ $\displaystyle p < x$. This is not a high school level problem, particularly for U.S. students, but it is not beyond approach. Let me put it this way for you, which should lead you to the answer:

What you're trying to avoid is that there is a prime $\displaystyle p, p > x$ such that [$\displaystyle 1 \equiv a \pmod p$], [$\displaystyle x \equiv b \pmod p$], $\displaystyle \dots$, [$\displaystyle x^n \equiv n \pmod p$] where [$\displaystyle a + b + \cdots n \equiv kp \equiv 0 \pmod p$]. So try looking at a few cases that solve the system of equations you're looking for them to solve and make a couple of hypotheses (although it should be moderately apparent, and if not, the OEIS is always a good tool, if it wouldn't feel too much like cheating. But don't only look at primes for values of $\displaystyle x$ or you will not find the sequence!). You will find that there is a certain class of numbers that admits the property which you seek, and a proof by contradiction follows without too much trouble. It is a much easier problem algebraically than when using strictly elementary tools, but I see how to go about it using only elementary techniques. Another hint: Ignore the '1' and try to avoid [$\displaystyle b + c + \cdots n \equiv p-1 \equiv -1 \pmod p$].

The proof is not short, but I will come back in a week or two and write it out for you if you still have not come up with the solution.