Is there a name for theorem ,$x^a - 1 \equiv 0 \mod {(x -1)}$, and how is my proof? I came to realise after practising some modular arithmetic that:
$x$, $(x-1)$$\space$ are$\space$ co-prime$ \space$ $\wedge$ $\space$ $x$, $(x-1)$ $\in$ $\mathbb{N}$ $\implies$ $x^\alpha \mod (x-1)$ = $1$ : $\forall$ $\alpha$ $\in$ $\mathbb{N}$
What is the name for this theorem, and who first proved it?
If someone could review the rigor of my proof that would be helpful:
Let the hypothesis be denoted as '$p$': $x$, $(x-1)$$\space$ are$\space$ co-prime$ \space$ $\wedge$ $\space$ $x$, $(x-1)$ $\in$ $\mathbb{N}$ 
Let the conclusion be denoted as '$q$': $\forall$ $\alpha$ $\in$ $\mathbb{N}$, $\space$ $x^\alpha \mod (x-1)$ = $1$ 
$p$ $\implies$ $q$.
Direct Proof:
Take the theorem $(A)^n \mod m$ $\iff$ $(A\mod m)^n \mod m$ : $A$ $\in$ $\mathbb{Z}$,$\space$ $n$, $m$ $\in$ $\mathbb{N}$: $m$ $\neq$ $0$
Then, given $p$, if we were to take the modulo $(x-1)$ of$\space$ $x^\alpha$ : $\alpha$ $\in$ $\mathbb{N}$,
$x^\alpha \mod (x-1)$ $\iff$ $(x \mod (x-1))^\alpha \mod (x-1)$
Since $x$ $-$ $(x-1)$ = $1$ (Given $p$):
$(x \mod (x-1))^\alpha \mod (x-1)$ $\equiv$ $[1]^\alpha \mod (x-1)$
We know that, $\forall$$\alpha$ , $1^\alpha = 1$
Hence, $[1]^\alpha\mod (x-1)$ $\equiv$ $[1] \mod (x-1)$
So,$\space$ $x^\alpha \mod (x-1)$ $\equiv$ $[1]$
Hence, $q$.
$p \implies q$. 
$QED$.
 A: Modulo $x-1$ clearly $x$ is $1$. So all the  powers of it  are also $1$.
A: First, let me compliment you on your enthusiasm and curiosity and ability to care about proofs.
Now the bad news.  I'm afraid your theorem is trivial.
$$x= (x-1)+1 \equiv 1 \mod (x-1)\\
⇒ x^{\alpha}\equiv 1^{\alpha} \equiv 1\mod (x-1).$$
A: Theorem: (reworded for  readability) $ \ $ For any two positive integers $x$ and $\alpha$:
$$
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).
$$
Q.E.D.
A: Who first proved it?
From Wikipedia: Book IX, Proposition 35 of Euclid's Elements expresses the partial sum of a geometric series in terms of members of the series. It is equivalent to the modern formula
$$
{x^\alpha-1\over x-1} = 1 + x + \ldots + x^{\alpha-1}.
$$
For a geometric series with integer terms, the partial sum $\displaystyle{x^\alpha-1\over x-1}$ is an integer. (Yet another restatement of your theorem.)
A: It doesn't have a name, I think. It is a consequence of $$x\equiv y\pmod z\implies \forall n\in \mathbb N\;[\;x^n\equiv y^n\pmod z\;]$$ which is  shown by induction on $n$, by using the basic result that $$[x\equiv y \pmod z\land x'\equiv y' \pmod z]\implies xx'\equiv yy'\pmod z.$$
