AR(1) Finding $\gamma_l$

I have $$\gamma_l = Cov(r_t, r_{t-l})$$ as a definition in my notes and now I need to find $$gamma_l$$ for a series $$r_t- m = p(r_{t-1} - m) + a_t$$ where $$r_t$$ is a linear time series with expected value $$m$$ and $$a_t$$ is a white noise.

Here is my attempt: $$Cov(r_t - m, r_{t-l} - m) = E[(r_t - m)(r_{t-l} -m )] = E[p^2(r_{t-1} - m)(r_{t-l-1} -m ) + pa_t(r_{t-l-1} -m ) + pa_{t-l}(r_{t-1} - m) + a_ta_{t-l}] = p^2E[(r_{t-1} - m)(r_{t-l-1} -m )] + 0 + pE[a_{t-l}(r_{t-1} - m)] + E[a_t]E[a_{t-l}]$$

Now if $$l=0, = p^2E[(r_{t-1} - m)(r_{t-l-1} -m )] + pE[a_{t-l}(r_{t-1} - m)] + s_a^2$$

Otherwise, $$= p^2E[(r_{t-1} - m)(r_{t-l-1} -m )] + pE[a_{t-l}(r_{t-1} - m)]$$.

In either case, I'm stuck here. I am supposed to get everything in terms of $$\gamma_{l-1}$$ (or $$\gamma_1$$ in the case of 0, but I am not sure how to proceed.

Let $$v_t=r_t-m$$, then $$v_t=pv_{t-1}+a_t$$. Covariance is invariant under de-meaning: $$\gamma_l=cov(v_t,v_{t-l})$$. Let’s first solve $$\gamma_0$$:

$$\gamma_0=cov(pv_{t-1}+a_t, pv_{t-1}+a_t)=p^2\gamma_0+\sigma_a^2 \rightarrow \gamma_0=\frac{\sigma_a^2}{1-p^2}$$

Note that $$cov(v_{t-1},a_t)=0$$ because previous output is irrelevant of the current white noise.

For $$l=1$$: $$cov(v_t,v_{t-1})=cov(pv_{t-1}+a_t,v_{t-1})=pcov(v_{t-1},v_{t-1})=p\gamma_0$$

For $$l=2$$: $$cov(v_t,v_{t-2})=cov(pv_{t-1}+a_t,v_{t-2})=pcov(v_{t-1},v_{t-2})=p\gamma_1=p^2\gamma_0$$

If you go on like this, you’ll have $$\gamma_l=p^l\gamma_0$$. Since $$\gamma_l=\gamma_{-l}$$, by definiton, this formula generalizes into the following:

$$\gamma_l=p^{|l|}\frac{\sigma_a^2}{1-p^2}$$