Number of elements which are cubes/higher powers in a finite field. This question is a slight generalization of This Question.
How many elements are there in a finite field of order $q$ which are : 


*

*Squares.

*Cubes.

*Higher powers.


I mean :
How many elements are there which are squares... 
How many elements are there which are cubes... 
how many elements are there which are higher powers... 
What all I could see is  :
For finding elements which are cubes I would consider :
$\eta : F^*\rightarrow F^*$ sending $x\rightarrow x^3$
For this If I know the kernel of $\eta$ then I would have $F^*/Ker(\eta) \cong \{x^3 : x\in F^*\}$
Which says that :
No of elements which are cubes are $\dfrac{|q-1|}{|Ker (\eta)|}$.
Now, Kernel of $\eta$ would be $\{x\in F^* : x^3=1\}$
All I know is if there is some subgroup (I am mentioning $\{1,x,x^2\}$ for $\{x\in F^* : x^3=1\}$)
Then It would be unique as a finite cyclic group can not have two subgroups of same order.
So, Now the problem is how to see for existence as we are through with uniqueness.
Suppose I prove uniqueness then I would say :
Number of elements of $F^*$ which are squares are :


*

*$\dfrac{q-1}{3}$ If $F^*$ have an element of order $3$ 

*$q-1$ If $F^*$ have no  element of order $3$ 


(I believe this would be totally dependent of nature of $q$)
Now the question is how do i make sure of existence with given nature of $q$.
I would be Thankful If some one can help me to see this and I would be happy to see further generalization.

How many elements of order $n$ are there in a Finite field of cardinality $q$

Thank you :) 
 A: This problem becomes easy if you know the theorem that the multiplicative group of a finite field is cyclic.
Here is a thread on the topic.
Finite subgroups of the multiplicative group of a field are cyclic
Edit:
The multiplicative group of the field with $q$ elements is therefore isomorphic to the additive group $\mathbf{Z}/(q-1)\mathbf{Z}$. There is an element of order 3 in this group if and only if $3 | (q-1)$.
Generally, if $n$ doesn't divide $q-1$, there will be no elements of order $n$ (but there might be some with order dividing $n$). If $n$ divides $q-1$, then $\mathbf{Z}/(q-1)\mathbf{Z}$ contains a unique subgroup of cardinal $n$, and there will be $\phi(n)$ elements of order $n$. 
A: The number of elements that are $n$-th powers is
$$
\frac{q-1} {\gcd(n,q-1)}. 
$$
Here the denominator is the size of your kernel. 
A: Lemma 1: $o(a^n)=\frac{o(a)}{\gcd(o(a),n)}$
Proof: Let $x=o(a), y=o(a^n), d=\gcd(x,n)$. We have $d|x$ and $d|n$, so $x=dv, n=du$. Then, $$\frac{o(a)}{\gcd(o(a),n)}=\frac{x}{d}=\frac{dv}{d}=v.$$ Since $x=o(a)$, we have $a^x=a^{dv}=1$ in $K$. Thus, $$(a^n)^v=(a^{du})^v=a^{duv}=(a^{dv})^u=1.$$ Thus, $y|v$. Then, $(a^n)^y=a^{ny}=1$, so $x|ny$. Thus, $$\frac{ny}{x}=\frac{duy}{dv}=\frac{uy}{v}\in\mathbf{Z}^+.$$ Now, by Bezout's Lemma, there exist $r,s$ such that $$rx+ns=d\implies rdv+dus=d\implies rv+us=1,$$ so $u$ and $v$ are coprime. Since $v|uy$ and $\gcd(u,v)=1$ we have $v|y$. However, we showed that $y|v$, so $v=y$. Finally, $x=dy$, so $o(a)\gcd(o(a),n)o(a^n)$, as desired. $\Box$
Suppose $K$ is a finite field of order $q$. Since $K^{\times}$ is cyclic, we can pick a generator $g$ in this group. Now, we have $$\text{number of dth powers}=\underbrace{\operatorname{ord}_{q}(g^d)=\frac{\operatorname{ord}_q(g)}{\gcd(\operatorname{ord}_q(g), d)}}_{\text{By Lemma 1}}=\frac{q-1}{\gcd(q-1, d)}.$$
