Question: I started with the claim that: If $n$ is even, $n^k$ is even.

But I have only gotten this far:
$$n = 2a,$$
$$n^k = (2a)^k,$$
$$n^k = 2^k a^k.$$

I am trying to reach the format where I can show that
$$n^k = 2a,$$
so that $n^k$ is even.

Answer: 
Consider the map from $S\_3$ to $S\_6$ which sends:

$$Id \to Id$$

$$(1,2)\to (1,4)(2,5)(3,6)$$

$$(1,3)\to (1,5)(2,6)(3,4)$$

$$(2,3)\to (1,6)(2,4)(3,5)$$

$$(1,2,3) \to (1,3,2)(4,5,6)$$

$$(1,3,2)\to (1,2,3)(4,6,5)$$

This is a homomorphism given by the Cayley theorem. The corresponding action is transitive.

Note: when we multiply permutations $ab$ the permutation $a$ acts first.
A subgroup of order $6$ of $S\_6$, say $\Sigma$, which acts transitively on the set $X:=\{1,\dots,6\}$, must have all the stabilizers trivial (Orbit-Stabilizer Theorem): $ \forall i\in X, \operatorname{Stab}(i)=\{Id\_X\}$, or equivalently: $ \forall i\in X, \sigma\in \Sigma\mid\sigma(i)=i\Rightarrow \sigma=Id\_X$. Therefore, we need to look at the elements of $S\_6$ which *do not fix any* element of $X$, namely:

1. $6$-cycles;
2. products of a $2$-cycle with a disjoint $4$-cycle;
3. products of $2$ disjoint $3$-cycles;
4. products of $3$ disjoint $2$-cycles.

Now, narrowing the survey to $\Sigma$'s isomorphic to $S\_3$, options 1 and 2 are ruled out because $S\_3$ is made of (further than the identity) $3$ elements of order $2$ and $2$ elements of order $3$. Therefore, any closed $\Sigma\subseteq S\_6$ made of the identity, three products of $3$ disjoint $2$-cycles, and two products of $2$ disjoint $3$-cycles, is a transitive subgroup isomorphic to $S\_3$.