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][1] 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)][2], 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 $\pi_6 O(7)=0$. 

  [1]: http://en.wikipedia.org/wiki/Homotopy_group
  [2]: http://felix.physics.sunysb.edu/~abanov/Teaching/Spring2009/Notes/abanov-cpA1-upload.pdf