It takes a long, painful, but straightforward calculation to see that the commutators of grade one elements $[e_i, e_j]$ of a Clifford algebra $\mathrm{Cl}(p,q)$ have exactly the same commutation relations as the rotations in the $(ij)$ plane that form the basis of $\mathfrak{so}(p,q)$. There is a different path to this result that I have yet to understand thoroughly, but on the surface it seems to involve a similarly unilluminating calculation to prove that $\mathrm{ad}_{[e_i, e_j]}$ maps vectors into vectors and (linear combinations of) bivectors into bivectors.
Is there a way to understand why it couldn’t be otherwise? (Not necessarily a complete proof, but at least some motivation to carry out the calculation knowing what to expect.)