# 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 $p$):

$(x \mod (x-1))^\alpha \mod (x-1)$ $\equiv$ $[1]^\alpha \mod (x-1)$