How to prove $\,x^a-1 \mid x^b-1 \iff a\mid b$

Hint $\rm\,\ mod\ \ x^{\large A}\!-1\!:\ \ \color{#c00}{x^{\large A}\equiv 1},\ \ \, so\ \ \ \smash[b]{\underbrace{x^{\large B}\equiv x^{\large B\ mod\ A}}} \equiv 1 \!\iff\! B\ mod\ A = 0 \!\iff\! A\mid B$

$\text{since by division}\ \ \rm B = AQ+R\,\Rightarrow\, x^{\large B}\equiv (\color{#c00}{x^{\large A }})^{\large Q} x^{\large R}\equiv {\color{#c00}1}^{\large Q} x^{\large R}\equiv x^{\large R},\ $ for $\rm\, R = B\bmod A$

The method in the above proof is called modular order reduction. It works in any ring (or monoid) since it uses only ring (or monoid) laws and consequent congruence product & power rules.

Remark $\ $ One can show much more. The polynomial sequence $\rm\ f_n = (x^n-1)/(x-1),\, $ like the Fibonacci sequence, is a strong divisibility sequence $\rm\: (f_m,f_n)\: =\: f_{\:(m,n)},\,$ where $\rm\,(a,b):=\gcd(a,b).\,$ The proof is simple - essentially the same as the proof of the Bezout identity for integers - see here. Thus we can view the polynomial Bezout identity as a q-analog of the integer Bezout identity, e.g. compare the Bezout identity for the gcd $\rm\ \color{#90f}3\, =\, (\color{#0a0}{15},\,\color{#c00}{21})\ $ in polynomial and integer form:

$$\rm \color{#90f}{\frac{x^3-1}{x-1}}\ =\ (x^{15} + x^9 + 1)\ \color{#0a0}{\frac{x^{15}-1}{x-1}}\ -\ (x^9+x^3)\ \color{#c00}{\frac{x^{21}-1}{x-1}}$$

for $\rm\ x = 1\ $ this specializes to $\ \ \color{#90f}3\ =\ (3)\ \color{#0a0}{15}\ -\ (2)\ \color{#c00}{21}.\, $ It is well-worth mastering these binomial divisibility properties since they occur quite frequently in number theoretical applications and, moreover, they provide excellent motivation for the more general study of divisibility theory, $ $ esp. in divisor theory form. For an introduction see Borovich and Shafarevich: Number Theory.