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.
Label $S\_3=\{\sigma\_1,\cdots,\sigma\_6 \}$. Each $\sigma\_i$ determines a permutation of $S\_3$ by $l\_i : \sigma\_j \mapsto \sigma\_i \sigma\_j$. This gives a one-to-one homomorphism $S\_3 \hookrightarrow S\_6$.

$S\_6$ acts on $S\_3$ by $\tau \cdot \sigma\_j=\sigma\_{\tau(j)}$. Each $\sigma\_i \in S\_3$ defines a permutation $\tau\_i$ of $\{1,\cdots,6\}$.

Moreover, for any $\sigma\_j$ and $\sigma\_k$ in $S\_3$, taking $\sigma=\sigma\_k\sigma\_j^{-1}$ one has $\sigma \sigma\_j=\sigma\_k$, that is the action of $S\_3$ (as a subgroup of $S\_6$) is transitive on $\{ \sigma\_1,\cdots,\sigma\_6 \}$.