The tangent bundle to a smooth structure on $S^7$ is classified by a map $S^7 \to G_7(R^{\infty})$. By the exact sequence for a fibration for the fiber bundle $O(7)\to V_7(R^\infty)\to G_7(R^\infty)$, we see that $\pi_7(G_7(R^\infty)) = \pi_6(O(7))$. But $\pi_6(O(7))=0$ (I found a table A1.1.3.2 of homotopy groups of orthogonal groups here(pdf), since this isn't in the stable range of Bott periodicity), so the tangent bundle is trivial, i.e. parallelizable.

The tangent bundle to a smooth structure on $S^7$ is classified by a map $S^7 \to G_7(R^{\infty})$. By the exact sequence for a fibration for the fiber bundle $O(7)\to V_7(R^\infty)\to G_7(R^\infty)$, we see that $\pi_7(G_7(R^\infty)) = \pi_6(O(7))$. But $\pi_6(O(7))=0$ (I found a table A1.1.3.2 of homotopy groups of orthogonal groups here(pdf), since this isn't in the stable range of Bott periodicity), so the tangent bundle is trivial, i.e. parallelizable.

Addendum: From the fibration $O(7)\to O(8)\to S^7$, we have the fibration sequence $$\pi_7(O(7))\to \pi_7(O(8))\to \pi_7(S^7)\to \pi_6 O(7)\to \pi_6 O(8)\to \pi_6 S^7=0.$$

Since $S^7$ is parallelizable (as may be shown via the octonians for example), there is a splitting $\pi_7(S^7)\to \pi_7(O(8))$. Hence we have an isomorphism $\pi_6 O(7)\to \pi_6 O(8) = \pi_6 O(\infty)$, since this is in the stable range. By Bott periodicity, $\pi_6 O(\infty)=0$, so $\pi_6 O(7)=0$.