To show that $2$ | $x^{2n} - x^n$, $n \in \mathbb{Z^+}$ is the following adequate? 
Show that $2$ | $x^{2n} - x^n$, $n \in \mathbb{Z^+}$

Let $x^n = a$ then $\frac{a}{2}$ $\implies a = 2q + r$ such that $r = 0 \vee r = 1$. 
So $x^{2n} = (x^n)^2 = a^2 = (2q + r)^2$, which means that $2|[x^{2n} - x^n ]= 2| [(2q + r)^2 - (2q + r)]$. If $r = 1$ we get $(4q^2 + 4q + 1) - 2q - 1 = 4q^2 + 2q = 2(2q^2 +1)$. Thus $2$|$x^{2n} - x^n$, $n \in \mathbb{Z^+}$. Is this reasoning correct? Is it adequate?

Your answer is OK, but it seems like an overkill. Just factor $x^{2n} - x^n$ into $x^n(x^n - 1)$ and use the fact that exactly one of two consecutive integers is even.
