Which assumption is stronger (p-BDHI) and DL? (p-BDHI): p-Bilinear Diffie-Hellman inversion problem

Given $P,sP,s^2P,...,s^pP$. Find $e(P,P)^{\frac{1}{s}}$.

(DL): discrete logarithm problem

Given $P,sP$. Find $s$.

To break (p-BDHI), attacker needs to break DL.

I think the two assumptions have same-level hard problem.

No one is stronger than another. Is it right?
p-BDHI is clearly at least stronger than DL: if you can break the DL problem, you can recover $s$, and then compute $e(P,P)^{1/s}$ (if you are in a group where inversion can be computed efficiently, which I think is always the case in known pairing groups).

On the other end, it is not clear that breaking p-BDHI could allow breaking the discrete logarithm problem, as, intuitively, breaking it only involve computing a group element and does not directly give a way to recover an exponent (this is informal as it could still be that p-BDHI allows too break the discrete logarithm problem, it does just seem far from obvious).