Change of variables and Galois Group 
Prove that the Galois group of $x^{5m}+5x^{5n}+5, \quad m>n$ is not $S_{5m}$.

I guess the 5 factors are given so that it will be related to the cyclotomic field, as if we are going to make a change of variable(?) $y=x^5$ and get $y^m+5y^n+5$ which has Galois group smaller than $S_{m}$, and then $y=x^5$ contributes maybe with $\mathbb{Z}/5\mathbb{Z}$ so that we get that the biggest group the Galois group can be is $S_{m}\times\mathbb{Z}/5\mathbb{Z}$ which is smaller than $S_{5m}$. However, I know this is really hand-wavy and I have no idea at this point. Thank you very much for your help!
 A: $\newcommand{\gal}{\textrm{Gal}}$
$\newcommand{\mq}{\mathbb Q}$ 
Let the splitting field of $p(x)=x^{5m}+5x^{5n}+5$ over $\mathbb Q$ be $E$ and the splitting field of $q(x)=x^m+5x^n+5$ over $\mathbb Q$ be $K$.  
The automorphism group we care about is that of $E$ over $\mq$. Since every root of $q(x)$ is a $5$th power of a root of $p(x)$, we have $\mq\leq K\leq E$. 
A theorem from Galois theory states that, in general, if $F$ is a field of nonzero characteristic and $K$ and $E$ are Galois extensions of $F$ s.t. $F\leq K\leq E$, then $\gal(E:F)/\gal(K:F)\cong\gal(E:K)$. 
So in this case, we have $\gal(E:\mq)/\gal(K:\mq)\cong\gal(E:K)$ which means $|\gal(E:\mq)|/|\gal(K:\mq)|=|\gal(E:K)|=[E:K]$.  
There are $m$ roots that generate $K$, and each root generating $E$ is a $5$th root of one of these. So, via the tower theorem, we can argue that $[E:K]\leq 20^m$ (I'll explain more at the bottom).  
This gives us $|\gal(E:\mq)|\leq |\gal(K:\mq)|20^m\leq 20^m m!$.  
You can easily show by weak induction that the above term is strictly less than $(5m)!$ whenever $m>0$.
So $|\gal(E:\mq)|<(5m)!$ which necessitates $\gal(E:\mq)\neq S_{5m}$.  

Tower Theorem Point: 
Consider a simpler case where $m=1$. In this case, if $r$ is the only root of $q(x)$, then $K$ will equal $\mq(r)$ and $E$ will equal $\mq(z_1\sqrt[5]r,z_2\sqrt[5]r,z_3\sqrt[5]r,z_4\sqrt[5]r,z_5\sqrt[5]r)$ where the $z_i$s are all the $5$th roots of unity.  
A basic facts about roots of unity is that all $n$th roots of unity can be expressed as a power of single $n$th root of unity, known as a primitive $n$th root of unity. Primitive $n$th are not unique, but they always exist.  
If $w$ is a primitive $5$th root of unity, then we can rewrite $E$ as the extension $\mq(\sqrt[5]r,w\sqrt[5]r,w^2\sqrt[5]r,w^3\sqrt[5]r,w^4\sqrt[5]r)$ which is clearly just equal to $\mq(w,\sqrt[5]r)$. So we have $[E:K]=[\mq(w,\sqrt[5]r):\mq(r)]=[\mq(w,\sqrt[5]r):\mq(\sqrt[5]r)][\mq(\sqrt[5]r):\mq(r)]$.  
Since $\sqrt[5]r$ is a root of $x^5-r\in \mq(r)[X]$, $[\mq(\sqrt[5]r):\mq(r)]\leq 5$. And, similarly, $w$ is a root of $\frac{x^5-1}{x-1}=x^4+x^3+x^2+x+1\in \mq[X]\subset \mq(\sqrt[5]r)[X]$, $[\mq(w,\sqrt[5]r):\mq(r)]\leq 4$.  
Putting all this together, we have $[E:K]\leq 4\cdot 5=20$.  
Using the tower theorem in the general case of $m$ roots of $q(x)$ just has you multiplying all these $20$s together $m$ times (as an upper bound). 
If you aren't familiar with the concept of primitive roots, you can still get a bound (albeit much weaker) of $120$. Because, in the case of $q(x)$ having just one root, like before, $[E:K]=|\gal(E:K)|\leq |S_5|=5!=120$. So instead of an upper bound of $20^m m!$ you would get $120^m m!$. This is still less than $(5m)!$ for $m>1$ (again, this can be shown by induction), but it is equal to $(5\cdot 1)!=120$. So with this weaker bound, the case where $m=1$, i.e. $p(x)=x^5+5$ will need to be dealt with separately.  
(please comment or edit for any corrections)
