Show that $p-1$ is a factor of $p^{10}-1$ and also of $p^{11}-1$.

(N/A) According to the Factor Theorem,if $(x-a)$ is a factor of a polynomial $f(x)$,then $f(a) = 0$.
For the polynomial $f(p) = p^{10}-1$,we check the factor $(p-1)$ by substituting $p=1$:
$f(1) = (1)^{10} - 1 = 1 - 1 = 0$.
Since the remainder is $0$,$(p-1)$ is a factor of $p^{10}-1$.
For the polynomial $g(p) = p^{11}-1$,we check the factor $(p-1)$ by substituting $p=1$:
$g(1) = (1)^{11} - 1 = 1 - 1 = 0$.
Since the remainder is $0$,$(p-1)$ is a factor of $p^{11}-1$.
Thus,it is proved that $(p-1)$ is a factor of both $p^{10}-1$ and $p^{11}-1$.