Is there a name for theorem ,$x^a - 1 \equiv 0 \mod {(x -1)}$, and how is my proof?

First, let me compliment you on your enthusiasm and curiosity and ability to care about proofs.

Now the bad news. I'm afraid your theorem is trivial.

$$x= (x-1)+1 \equiv 1 \mod (x-1)\\ ⇒ x^{\alpha}\equiv 1^{\alpha} \equiv 1\mod (x-1).$$

Theorem: (reworded for readability) $ \ $ For any two positive integers $x$ and $\alpha$: $$ x^\alpha - 1 \quad\mbox{ is divisible by }\quad x - 1. $$

Proof: The difference $x^\alpha - 1$ can be factored as $$ x^\alpha - 1 = (x-1)(x^{\alpha-1}+x^{\alpha-2}+\ldots+x+1). $$ Q.E.D.

Who first proved it?

From Wikipedia: Book IX, Proposition 35 of Euclid's Elements expresses the partial sum of a geometric series in terms of members of the series. It is equivalent to the modern formula $$ {x^\alpha-1\over x-1} = 1 + x + \ldots + x^{\alpha-1}. $$ For a geometric series with integer terms, the partial sum $\displaystyle{x^\alpha-1\over x-1}$ is an integer. (Yet another restatement of your theorem.)