Q:

Is it true that $a^{k(p-1)+b} \;\stackrel{p}{\equiv} \;\;a^b\;$?

$$a^{k(p-1)+b} \;\stackrel{p}{\equiv} \;\;a^b\;?$$
$p$ prime number and $a,b,k\in\mathbb{N}^+$. And $p$ does not divide $a$.  
According to Fermat's Little theorem $a^{p-1}\stackrel{p}{\equiv}1$. So $a^{k(p-1)} \stackrel{p}{\equiv}1$ by splitting into k parts and apply modulus independently we get $1^k = 1$. Am I missing something?

A:

As $(a,p)=1$  using Fermat's Little Theorem $a^{p-1}\equiv1\pmod p$
So, $a^{k(p-1)}\equiv1^k\equiv1\pmod p$
In fact, if $a^x\equiv a^y\pmod p$ where $(a,p)=1$ and integers $y, x(>y)$
$\implies a^{x-y}\equiv1\pmod p$
We can prove if $d=ord_pa, d$ divides $(x-y)$
As $a^{p-1}\equiv 1\pmod p,$  so $ord_pa$ divides $p-1$ 
If $(p-1)$ divides $(x-y), d$ will divide $(x-y)$

