Is this matrix obviously positive definite?

It is positive-definite for all $n$.

The way I see it is by noting that it has Cholesky decomposition $$A = LL^\top$$ where $$L = \begin{bmatrix} 1 & & & \\a & \sqrt{1-a^2} & & &\\a^2 & a\sqrt{1-a^2} & \sqrt{1-a^2} & &\\a^3 & a^2\sqrt{1-a^2} & a\sqrt{1-a^2} & \sqrt{1-a^2} &\\\vdots & \vdots & \vdots & \vdots & \ddots\end{bmatrix}.$$

It's not hard (though not "obvious", perhaps) to see that this decomposition is correct: the dot product of a row with itself telescopes to 1, and of two different rows, to the appropriate power of $a$.

Is this matrix always positive definite? If so, what is the simplest way to see that?

It is. A simple way to prove it is to note that this is (a submatrix of) the covariance matrix of any process $(x_n)$ such that, for every integer $n$, $$x_n=ax_{n-1}+z_n$$ where $(z_n)$ is i.i.d. with variance $1-a^2$. An alternative formulation of this recursion is that, for every $n$, $$ x_n=\sum_{k\geqslant0}a^kz_{n-k} $$ When one computes the covariance of $x_n$ and $x_{n+m}$, every product $z_iz_j$ with $i\ne j$ disappears because the $z_k$ are independent, and every product $z_i^2$ yields $1-a^2$, hence $$ \mathrm{Cov}(x_n,x_{n+m})=\sum_{k,\ell\geqslant0}(1-a^2)a^{k+\ell}\cdot[n-k=n+m-\ell], $$ that is, $$ \mathrm{Cov}(x_n,x_{n+m})=(1-a^2)\sum_{k,\ell\geqslant0}a^{k+\ell}\cdot[\ell=m+k]=(1-a^2)\sum_{k\geqslant0}a^{m+2k}=a^m. $$ In other words, the infinite dimensional vectors $\mathbb x=(x_n)$ and $\mathbb z=(z_n)$ are such that $\mathbb x=\mathbb T\cdot \mathbb z$ where $\mathbb T$ is the lower triangular infinite dimensional matrix defined by $T_{n,n+k}=0$ if $k\geqslant1$ and $T_{n,n-k}=a^k$ if $k\geqslant0$. The infinite dimensional matrix $\mathbb A$ whose every finite dimensional matrix $A$ is a submatrix, is $$ \mathbb A=\mathrm{Cov}(\mathbb T\mathbb z,\mathbb T\mathbb z)=\mathbb T\cdot\mathrm{Cov}(\mathbb z,\mathbb z)\cdot \mathbb T^*=\mathbb T\cdot \mathbb T^*. $$ In particular for every vector $y$ of size $n$, extending $y$ to an infinite dimensional vector $\mathbb y$ by adding zeroes, one gets $$ y^*\cdot A\cdot y=\mathbb y^*\cdot \mathbb A\cdot \mathbb y=\mathbb y^*\cdot \mathbb T\cdot \mathbb T^*\cdot \mathbb y=\|\mathbb T^*\cdot \mathbb y\|^2\geqslant0. $$

Quick checks in Maple show that for the first cases, the determinant of $A$ is $(1-a^2)^{n-1}$, which is positive. I think this can be proved recursively. This proves that all the diagonal submatrices have positive determinant, which means that $A$ is positive definite..