Prove that $ A^n := \{ a^n : a \in A \}$ and $A_n := \{ g: g \in A : g^n = 1 \} $ are characteristic subgroups of $A$. The Statement of the Problem:
Let $A$ be an abelian group and $n$ a positive integer. Prove that
$$ A^n := \{ a^n : a \in A \} $$
$$ \text{and} $$
$$ A_n := \{ g: g \in A : g^n = 1 \} $$
are characteristic subgroups of $A$.
Where I Am:
So, it looks like what I want to show is that $\forall \phi \in \text{Auto}(A), \phi (A^n) = A^n$. If I understand correctly, the case of inner automorphisms is clear:
$$ \phi _ a (a^n) = aa^na^{-1} = aa^{-1}a^n = a^n .$$
But, is there some way to represent an arbitrary automorphism and just "check" it? Or is this not as simple as I've hoped (or even simpler!)?
EDIT: I've got an idea for the first one, but I don't think it's correct...
$a^n = aaa...a$, i.e., the product of $n$ $a$'s. Now, because $\phi$ is a homomorphism:
$$ \phi (a^n) = \phi (aaa...a) = \phi(a)\phi(a)\phi(a)...\phi(a). $$
And because $\phi$ is bijective:
$$ \phi^{-1}[\phi(a)\phi(a)\phi(a)...\phi(a)] = aaa...a = a^n. $$ 
Q.E.D?
 A: I think it's easier (algebraically) than you're making it, but we have to think more if we want less algebra. I think you have the big idea, but I wouldn't say it's entirely airtight $\ddot\smile$. 
At least, I think the big idea of yours is that $\phi(a^n) \in A^n$, so $\phi(A^n) \subseteq A^n$. Mentioning this would be good: "Since $\phi(a) \in A$, we have that $\phi(a^n) = \phi(a)^n \in A^n$," for example.
I'm not sure you showed the reverse containment though, this is where the commentary is really important. To show that $A^n \subseteq \phi(A^n)$, show that every $a^n$ can be written as an image under $\phi$ of some $n$th power: $$a^n = \phi\big(\underbrace{\phi^{-1}(a)^n}_{\in A^n}\big) \in \phi(A^n).$$ But that was super confusing for me to figure out exactly what needed written down!

An alternative:
You already showed that $\phi(a^n) = \phi(a)^n$, and I would use that.
So, if $A^n = \{a^n : a \in A\}$, we have 
\begin{align*}
\phi(A^n) 
&= \{\phi(a^n) : a \in A\} \tag{definition of image} \\
&= \{\phi(a)^n: a \in A\} \tag{property of homomorphisms}\\
&= A^n
\end{align*}
where the latest equality is simply because $\phi: A \to A$ is a bijection; we have $$A = \{a: a \in A\} = \{\phi(a): a \in A\},$$
and so $\{\phi(a)^n : a \in A\}$ must be $\{a^n : a \in A\}$, just a bit "mixed up" perhaps (as $a$ runs through $A,\ \phi(a)$ does too).
