I would use the sectional curvature. Pick a point $p\in \mathbb S^{n-1}$ and a two-dimensional subspace $V\subset T_p S^{n-1}$. The geodesics starting from $p$ in directions lying in $V$ are all contained in the $3$-dimensional subspace $W\subset \mathbb R^n$ such that $T_p S^{n-1}\subset T_p W$. Therefore, the calculation of sectional curvature amounts to calculating the Gaussian curvature of $S^{n-1}\cap W$, which is just $S^2$. For that, there is a lot of explicit formulas which yield $1/r^2$ in reasonable time.
Once the sectional curvature is known to be $1/r^2$, the Ricci is determined from one version of the definition: average of sectional curvatures through a given vector (times $n-1$).
[Added] I used the definition of sectional curvature involving the exponential map, as in Helgason's Differential geometry. I quote: let $N_0$ be a normal neighborhood of $0$ in $T_pM$. Let $S$ be a two-dimensional subspace of $T_pM$. Then $\exp (N_0\cap S)$ is a 2-dimensional submanifold of $M$ with induced Riemannian structure. Its curvature at $p$ is called the sectional curvature of $M$ at $p$ along the plane section $S$.
Old-fashioned as it may be, this definition makes it as clear as possible where sectional comes from... Incidentally, Helgason defines the curvature of a 2-dimensional manifold by $K=\lim_{r\to 0} \frac{12}{r^2}\frac{A_0(r)-A(r)}{A_0(r)}$ where $A_0(r)$ and $A(r)$ stand for the areas of a disk $B_r(p)\subset T_pM$ and of its image under the exponential map. For the 2-sphere of radius $R$ we have $A(r)=2\pi R^2(1-\cos (r/R))=\pi r^2-\pi r^4/(12R^2)+o(r^4)$, hence $K=\lim_{r\to 0} \frac{12}{r^2}\frac{\pi r^4 /(12R^2)}{\pi r^2}=\frac{1}{R^2}$
Yeah... I like the exponential map and I hate tensors.