divisibility question in abstract algebra over a field $d|n \Rightarrow x^{p^d} - x$ divides $x^{p^n} - x$ over $\mathbb{F}_{p}$ where $\mathbb{F}$ is a field.
Attempt: 
$d|n \Rightarrow x^{p^d} - x$ divides $x^{p^n} - x \Rightarrow x^{p^d - 1} - 1$ divides $x^{p^n - 1} - 1 \Rightarrow p^d - 1$ divides $p^n - 1$ since $p^n - 1 = (p^d)^{\frac{n}{d}}-1$.
How does this look?
 A: You assume the result as your first step, and after that it's not clear what your goal is.  But you've come across an important fact, which is that what you're trying to prove is related to the fact that $p^d-1$ divides $p^n-1$.
First, let's make sure that we can prove this fact from scratch.  Let's write $n=kd$.  Then why does $p^d-1$ divide $p^{kd}-1$?  Here's a hint: if $y=p^d$, is it true that $y-1$ divides $y^k-1$?
And in fact, once we know that $p^d-1$ divides $p^n-1$, we can use the same trick to finish the problem.  If $p^n-1 = m(p^d-1)$, and z = $x^{p^d-1}$, then $x^{p^d-1} -1 = z-1$ and $x^{p^n-1} -1 = z^m-1$.
Do you see how to finish from here?
A: $$d\mid n\implies n= kd\;,\;\;\text{so}:$$
$$x^{p^n}-1=\left(x^{p^d}\right)^{p^{kd-d}}-1=\left(x^{p^d}-1\right)\left(\left(x^{p^d}\right)^{p^{kd-d}-1}+\left(x^{p^d}\right)^{p^{kd-d}-2}+\ldots+x^{p^d}+1\right)$$
A: You can either drop to the integer case by doing (not exactly) what you did (notice how your first line is the actual result...), which is fine, but there are some other tricks! You can use induction on $k = n/d$ as follows : 
$$
x^{p^n} - x = x^{p^{kd}} - x = x^{p^{kd}} - x^{p^{(k-1)d}} + x^{p^{(k-1)d}} - x.
$$
By induction you only need to prove that $x^{p^d} - x$ divides $x^{p^{kd}} - x^{p^{(k-1)d}}$. Then 
$$
x^{p^{kd}} - x^{p^{(k-1)d}} = \left(x^{p^{(k-1)d}}\right)^p - \left(x^{p^{(k-2)d}}\right)^p = (x^{p^{(k-1)d}} - x^{p^{(k-2)d}}) f_1(x)
$$
using the identity 
$$
a^n - b^n = (a-b)\left( \sum_{i=0}^{n-1} a_i b^{n-1-i} \right)
$$ 
(The sum here would correspond to $f_1(x)$.) You can see that the factor obtained has the same shape as the previous one except that $k$ is replaced by $k-1$, so you can keep going until $k$ becomes $1$ :
$$
x^{p^{kd}} - x^{p^{(k-1)d}} = \left(x^{p^{(k-1)d}} - x^{p^{(k-2)d}}\right)f_1(x) = \dots = (x^{p^d} - x)f_1(x) f_2(x) \dots f_{k-1}(x).
$$
More generally, you can show that if $(p(x),q(x))$ stands for the g.c.d. of those two polynomials (and/or integers), then
$$
(x^m - 1, x^n - 1) = x^{(m,n)} - 1.
$$
The proof is quick : if $m > n$, write $m = qn + r$. Then 
$$
x^m - 1 = x^m - x^{m-r} + x^{m-r} - x^{m-2r} + \dots x^{n+r} - x^r + x^r - 1 = \left( \sum_{j=1}^q x^{jn} - x^{(j-1)n} \right) + x^r - 1 
= (x^n - 1)\left( \sum_{j=1}^q x^{(j-1)n} - x^{(j-2)n} \right) + x^r-1.
$$
Therefore you can apply the Euclidean algorithm on $m$ and $n$. This gives you
$$
(x^{p^d} - x, x^{p^n} - x) = x (x^{p^d-1} - 1, x^{p^n-1} -1) = x (x^{(p^d-1,p^n-1)} - 1) = x^{(p^d-1,p^n-1) + 1} - x
$$
and so $x^{p^d} - x$ divides $x^{p^n} - x$ if and only if $(p^d-1,p^n-1) = p^d-1$, i.e. if and only if $p^d-1 | p^n -1$. There are many ways to prove that $p^d - 1 | p^n - 1$ if and only if $d | n$, which is a stronger statement than what you were trying to prove! The easiest way is probably to write $n = qd + r$ with $0 \le r < d$ and write
$$
p^n - 1 = p^n - p^r + p^r - 1 = p^r(p^qd - 1) + p^r - 1 = p^r (p^d-1)\left( \sum_{j=0}^{q-1} p^{qd} \right) + p^r - 1
$$
and since $r < d$, $p^d-1 | p^n-1$ if and only if $r=0$, i.e. if and only if $d | n$.
Hope that helps.
