Algebraic Closure of $\mathbf F_p$ [Lang, Algebra, Chapter 6, Problem 22] 
Problem. Let $K$ be the field obtained from $\mathbf F_p$ by adjoining all primitive $\ell$-th roots of unity for primes $\ell\neq p$.
  Then $K$ is algebraically closed.

It suffices to show that the polynomial $x^{p^n}-x$ splits in $K$ for all $n$.
In order to show this, it in turn suffices to show that the polynomial $x^{q^n}-1$ splits in $K$ for all primes $q\neq p$ and all $n$.
This is because $x^{p^n}-1= x(x^{p^n-1}-1)$.
Say $p^n-1=p_1^{k_1} \cdots p_m^{k_m}$, where $p_i$'s are distinct primes.
Assuming each $f_i(x):=x^{p_1^{k_i}}-1$ splits in $K$, we deduce that $K$ has a primitive $p_i^{k_i}$-th root of unity for all $1\leq i\leq m$ since each $f_i$ is separable by the derivative test. If $\zeta_i$ denotes the primitive $p_i^{k_i}$-th root of unity in $K$, then we see that $\zeta_1\times \cdots\times \zeta_m$ is a primitive $p_q^{k_1}\times \cdots \times p_m^{k_m}$-th root of unity and we see that $x^{p^n-1}-1$ splits in $K$.

So the problem boils down to showing that $x^{q^n}-1$ splits in $K$ for all primes $q\neq p$ and all $n$.

I am stuck here.
 A: The idea is that given any prime power $ q^k $, we may take a prime $ w $ such that $ w $ divides $ p^{q^k} - 1 $ but does not divide $ p^{q^{k-1}} - 1 $, in other words, such that $ p $ has order $ q^k $ modulo $ w $. First, assuming the existence of such a prime $ w $, we observe that $ \mathbf F_p(\zeta_w) $ is the finite field with $ p^{q^k} $ elements, so it is the splitting field of $ X^{p^{q^k}}  - X $ over $ \mathbf F_p $. Now, we proceed with the argument.
To see that such a prime $ w $ exists, we use the polynomial identity
$$ \frac{(1 + X)^q - 1}{X} = \sum_{k=0}^{q-1} C(q, k+1) X^{k} $$
and write
$$ a = \frac{p^{q^k} - 1}{p^{q^{k-1}} - 1} = \sum_{j=0}^{q-1} C(q, j+1) (p^{q^{k-1}} - 1)^j $$
Clearly, we have $ a > q $. On the other hand, if a prime $ v $ divides both $ a $ and the denominator, it must also divide $ q $ by the sum on the right hand side, and since $ q $ is prime we must have $ v = q $. However, in that case $ q^2 $ cannot divide $ a $, so $ a $ has a prime factor $ w \neq q $. Since $ w $ cannot be a divisor of the denominator, it is the desired prime number.
