Showing a certain time series is stationary I have been trying to solve the following problem, but have not been successful yet. I was hoping anyone could nudge me in the correct direction.

Let $\{ w_t; t = 0,1,\ldots \}$ be a white noice process with variance $\sigma^2$ and let $|\phi| < 1$ be a constant. Consider $x_0 = \frac{w_o}{\sqrt{1-\phi^2}}$ and
$$
x_t = \phi x_{t-1} + w_t,\ \ \ \ t=1,2,\ldots.
$$
Show that $var(x_t)$ is constant in $t$.


MY ATTEMPT:
Firstly, by recursively writing out $x_t$ it can be seen that:
$$
x_t = \phi^t x_0 + \sum_{k=0}^t \phi^k w_{t-k} = \phi^t \frac{w_0}{\sqrt{1-\phi^2}} + \sum_{k=0}^t \phi^k w_{t-k}.
$$
By linearity of the expectation, it follows that $E(x_t) = 0$. Thus $var(x_t) = E(x_t^2)$.
\begin{align}
var(x_t) &= E(x_t^2) = E([\phi^t \frac{w_0}{\sqrt{1-\phi^2}} + \sum_{k=0}^t \phi^k w_{t-k}]^2) \\
&= \phi^{2t}\frac{1}{1-\phi^2} E(w_0^2) + 2 \frac{\phi^t}{\sqrt{1-\phi^2}}E(w_0\sum_{k=0}^t \phi^k w_{t-k}) + E([\sum_{k=0}^t \phi^k w_{t-k}]^2) \\
&= \phi^{2t}\frac{\sigma^2}{1-\phi^2} + 2 \frac{\phi^{2t}}{\sqrt{1-\phi^2}}\sigma^2 + \sigma^2 \sum_{k=0}^t \phi^{2k} \\
&= \phi^{2t}\frac{\sigma^2}{1-\phi^2} + 2 \frac{\phi^{2t}}{\sqrt{1-\phi^2}}\sigma^2 + \sigma^2 \frac{1-\phi^{2(t+1)}}{1-\phi^2} \\
&= \frac{\sigma^2}{1-\phi^2}( \phi^{2t} + 1 - \phi^{2t} \phi^2) + 2 \frac{\phi^{2t}}{\sqrt{1-\phi^2}}\sigma^2 \\
&= \frac{\sigma^2}{1-\phi^2} + \frac{\sigma^2}{1-\phi^2}(\phi^{2t}(1-\phi^2)) + 2 \frac{\phi^{2t}}{\sqrt{1-\phi^2}}\sigma^2 \\
&= \frac{\sigma^2}{1-\phi^2} + \sigma^2 \phi^{2t} + 2 \frac{\phi^{2t}}{\sqrt{1-\phi^2}}\sigma^2.
\end{align}
This is where I am stuck: I believe the second and third term should somehow cause the dependence on $t$ to cancel out, but can't seem to figure it out. Is this the way to go or am I missing something else? In any case, thanks for your time!
 A: Trivially it can be seen that $\text{Var}(x_0)$ is constant.
Observe
$$\begin{align}
&x_1 = \dfrac{\phi w_0}{\sqrt{1 - \phi^2}} + w_1 \\
&x_2 = \dfrac{\phi^2w_0}{\sqrt{ 1- \phi^2}} + \phi w_1 + w_2 \\
&x_3 = \dfrac{\phi^3w_0}{\sqrt{1 - \phi^2}} + \phi^2 w_1 + \phi w_2 + w_3 \\
&\vdots \\
&x_t = \dfrac{\phi^tw_0}{\sqrt{ 1- \phi^2}} + \sum_{k=1}^{t}\phi^{t-k} w_k\text{.}
\end{align}$$
Using the fact that $\{w_t: t \geq 0\}$ are uncorrelated random variables, we have that their pairwise covariances are $0$, thus
$$\text{Var}(x_t) = \dfrac{\phi^{2t}\sigma^2}{1-\phi^2} + \sum_{k=1}^{t}\phi^{2(t-k)}\sigma^2 = \sigma^2\left(\dfrac{\phi^{2t}}{1-\phi^2} + \phi^{2t}\sum_{k=1}^{t} \phi^{-2k} \right)\text{.}$$
We have that
$$\begin{align}
\sum_{k=1}^{t} \phi^{-2k} &= \phi^{-2}+\phi^{-4}+\cdots+\phi^{-2t} \\
&= \phi^{-2}\left[1+\phi^{-2}+\cdots+\phi^{-2(t-1)}\right]  \\
&= \phi^{-2}\left(\dfrac{1-\phi^{-2t}}{1-\phi^{-2}} \right) \\
&= \dfrac{1-\phi^{-2t}}{\phi^2 - 1} \\
&= \dfrac{\phi^{-2t} - 1}{1-\phi^2}
\end{align}$$
thus
$$\phi^{2t}\sum_{k=1}^{t}\phi^{-2k} = \dfrac{1 - \phi^{2t}}{1-\phi^2}$$
and
$$\text{Var}(x_t) = \dfrac{\sigma^2}{1-\phi^2}\text{.}$$
