How would you prove that the altitudes of an obtuse angled triangle always intersect outside of it?

I believe an indirect proof would be your best bet. Begin by supposing that the orthocenter of an obtuse triangle is on the interior, and show that that condition leads to a contradiction.

In ∆ ABC, let ∠ABC > 90°.

Suppose orthocenter K lies on the triangle interior.

Produce AK to meet side BC at F, the foot of the altitude from A. Since K is in the interior of ∠BAC, so must be F. Point F must lie between B and C. It follows then that ∠ABC and ∠ABF are one and the same.

By definition of an altitude, ∠AFB = 90°.

That makes ∆AFB a right triangle, so ∠ABF is acute, and ∠ABC is acute. This conflicts with the given condition that ∠ABC is obtuse.

Because of this contradiction, we must reject the proposition that the orthocenter is on the interior of the triangle.

I still have not proved that the orthocenter cannot lie on the triangle perimeter. I will leave the cleanup to you.

The altitudes to either of the sides of the obtuse angle are exterior to the triangle, hence any intersection of them or with anything else must be exterior to the triangle.

Obviously because one of the altitudes of an obtuse triangle

is outside the triangle.