Prove that for any $n > 0$, if $a^n$ is even, then $a$ is even the proof at hand is:

Prove that for any $n > 0$, if $a^n$ is even, then $a$ is even.
Hint: Contradiction

So I know to start the problem in a contradiction format would be:
$a^n$ is even, then $a$ is odd so that $a = 2k+1$.  Then plug in that into $a^n$, we get $(2k+1)^n$.  This is where I become stuck.
Thanks in advance!
 A: Hint: expand using the binomial theorem.
A: $$(2k+1)^n=2q+1\quad,\quad \text{for some }q\in\Bbb Z$$since$$(2k+1)^n{=\sum_{i=0}^n(2k)^i\\=1+\sum_{i=1}^n(2k)^i\\=1+2\cdot\sum_{i=1}^n2^{i-1}\cdot k^i\\=1+2q}$$
A: By Euclid's lemma, if $2|a^n$ then $2|a$ or $2|a$ or $2|a$ or ... or $2|a$,
so we conclude that $2|a$.
A: The contrapositive is that $$a \text{ odd }\implies a^n \text { odd}$$
Use induction here: if $a^1=2k_1+1$, then $a^2=(2k_1+1)^2=2(2k_1^2+2k_1)+1=2k_2+1$
hence true for $n=2$
Then assume true for $n=m$, $(2k_1+1)^m=2k_m+1$
Then $$(2k_1+1)^{m+1}=(2k_m+1)(2k_1+1)=4k_mk_1+2k_m+2k_1+1=2(2k_mk_1+k_m+k_1)+1$$
Hence true with $k_{m+1}=2k_mk_1+k_m+k_1$
A: You can prove this by induction: 
For $n=1$ the claim is clear. 
Assume that if $a^n $ is even then $a $ is even. 
Let $a^{n+1}$ be even. Then:
$a^{n+1} = a a^n$ .
If a is not even, then $a^n$ must be even because otherwise their product won't be even. 
But then , by the induction hypothesis , $a$ is even, contradiction. 
So $a$ must be even, as wanted.
A: Prove by contradiction : If $a^n$ is even then a is an odd number.
We have : a = 2k + 1
By assumption above : $(2k+1)^n$ = 2q for all q $\in$ Z
Using binomial theorem we have : $(2k+1)^n$ = $\sum_{i = 0}^n C_i^n(2k)^i$ = 1 + 2$\sum_{i = 1}^n C_i^n(2k)^{i - 1}$.
We can derive that q is a rational number contradicting the fact that q is an integer.
