Let $X$ be a projective variety (so, with some embedding) with the following curious property: for **every** hyperplane section $H$, we have that $X-H \cong \mathbb{A}^n$. Then is $X$ necessarily isomorphic to $\mathbb{P}^n$?

Assuming we are over the complex numbers for simplicity, one can show that $X$ has the same Hodge diamond as $\mathbb{CP}^n$ using purely topological arguments (such as long exact sequence of a pair, duality as in the proof of Lefschetz hyperplane, etc.) but this does not rule out these potential **fake** projective spaces.

This question is motivated by the fact that spheres are characterized by a similar property; i.e., a closed oriented manifold that is contractible upon removal of any point will be homeomorphic to a sphere.