Growing Ls

First thing to note is that as 2 can be tiled, so can all powers of 2.
By the same reasoning, if size n can be tiled, so can 2n.
So we only need to be able to show that all odd (well, all primes really) can be tiled, and we're done.

3 can be tiled:

from 3 we can tile 5, from 5 we can tile 7 and from 7 we can tile 9 (only half shown because of the symmetry)

To prove it can always be done, we just need to prove that for any odd n, the n+2th L can be tiled:

This means filling the inverted L with the dimensions given.
3 different cases: when n is congruent to 0, 1 or 2 mod 3.
When n congruent to 0, the L can be split into 3 n by 2 rectangles, which can be tiled (see 3 to 5 above)
When n is congruent to 1, n+2 is congruent to 3, so the L can be split into a 4 by (n+2) rectangle (tiled with 3 by 2 rectangles) and an (n-4) by 2 rectangle - (n-4) is congruent to 0 mod 3 so this can be tiled.
When n is congruent to 2, we split the shape into an (n-2) by 2 rectangle (can be tiled with 3 by 2 rectangles), an (n-2) by 4 rectangle (again can by tiled with 3 by 2 rectangles) and a size 2 L, which we know we can tile.