How can I prove whether a $9\times 9$ square can be filled with L-shaped pieces in a completely "regular" way?

Question

There are a great many ways to fill a $9\times 9$ square with L-shaped pieces. One of them is below.

Now, note that there are eleven $2\times 3$ rectangles that are formed, as well as a larger L shape. There is one "irregular" piece, which has been colored in green. What I think of as being "regular" is a bit subjective (as in aesthetically appealing), but I think it suffices to define a regular piece as being part of a rectangle or larger L-shape.

I conjecture that there must be at least one irregular piece. How can I prove this? (Alternatively, if I'm wrong, what would be a counterexample?)

The proof I'm thinking of is that all rectangles that can fit in a $9\times 9$ and can be constructed from L-pieces must have at least one side with even length, and that larger L-shapes also have dimensions that are of even lengths. Hence, as $9 \cdot 9 = 81$ is odd and all "regular" formations have even numbers of squares in them, there must be at least one square that does not fit into a "regular" formation, which them requires at least one "irregular" L-piece, which completes the proof. Is this rigorous enough (after adding mini-proofs that show why regular shapes must have even numbers of squares in them), or am I lacking important details?

Answer 1

Actually if you consider a "larger L shape" as regular, you can take your green L as part of such a larger L shape:

Answer 2

The $9 \times 9$ square takes $27$ L pieces to cover it. If the only regular pieces are $2 \times 3$ rectangles (which take two pieces) and your larger L (which takes four pieces), there must be at least one odd L piece because both the regular shapes take an even number of pieces.