Apply mathematical induction on $n$.

When $n=1$. Because $k$ is odd, therefore $k\equiv \pm 1, \pm 3\pmod{8}\implies k^2 \equiv 1\pmod{8}$. The claim holds.

Assume the claim hold when $n = m \ge 1$, i.e. $k^{2^m}\equiv 1\pmod{2^{m+2}}$. Let $k^{2^m} -1 = 2^{m+2}\cdot t$ where $t$ is an integer.

Then when $n=m+1$, we have $$\begin{array}{l} k^{2^{m+1}} -1 = \left(k^{2^m}\right)^2 - 1=\left(2^{m+2}\cdot t + 1\right)^2-1\\ = 2^{2m+4}t^2 + 2^{m+3}t=2^{m+3}\left(2^{m+1}t^2+t\right)\end{array}$$

Therefore, $2^{m+3}\mid \left(k^{2^{m+1}} -1\right)$. Hence, $k^{2^{m+1}}\equiv 1\pmod{2^{m+3}}$.