Need help on continued farction

I'm trying to show the following argument but have a problem

If $\displaystyle \frac{p_k}{q_k}$ is the kth convergent of $\displaystyle [a_0;a_1,...,a_n]$, show that for $\displaystyle 0 \leq k\leq n$, $\displaystyle q_k \geq 2^{\frac {k-1}{2}}$

I use that fact that

$\displaystyle q_k = a_kq_{k-1}+q_{k-2} \geq 2q_{k-2} \geq 2$, and

$\displaystyle 2q_{k-2} = a_{k-2}q_{k-3}+q_{k-4} \geq 2^2q_{k-4} \geq 2^2$ etc,

we can see that $\displaystyle q_k$ is decreasing in $\displaystyle 2n$ steps.

But why $\displaystyle q_k \geq 2^{\frac {k-1}{2}}$ instead of $\displaystyle 2^{\frac{k}{2}}$?

Can anyone explain it to me? Thank you.