Tiling of a deficient $7\times7$ chessboard with L trominoes

combinatoricstiling

Prove that a $$7\times7$$ chessboard with one square removed can always be tiled by $$L$$ trominoes.

I'm looking for a reasonably simple proof. I was able to prove some specific cases, For instance, when the central square is deleted, the chessboard can be partitioned into four $$4\times3$$ rectangles, which can be easily tiled.

However, I was unable to prove the general case. Any help would be greatly appreciated.

$$\begin{array}{|c|c|c|c|c|c|} \hline X&X&\circ&\circ&\triangle&\triangle&\circ\\ \hline X&X&\circ&\square&\triangle&\circ&\circ\\ \hline \circ&\circ&\square&\square&\blacksquare &\blacksquare &\square\\ \hline \circ&\triangle&\circ&\circ&\blacksquare&\square&\square\\ \hline \triangle&\triangle&\circ&\square&\square&\triangle&\triangle\\ \hline \circ&\circ&\triangle&\square&\circ&\triangle&\square\\ \hline \circ&\triangle&\triangle&\circ&\circ&\square&\square\\ \hline \end{array}$$
$$\begin{array}{|c|c|c|c|c|c|} \hline \square&\square&\circ&\triangle&\triangle&\circ&\circ\\ \hline \square&\circ&\circ&\triangle&\blacksquare&\blacksquare&\circ\\ \hline X&X&\square&\square&\blacksquare &\square &\square\\ \hline X&X&\circ&\square&\triangle&\triangle&\square\\ \hline \triangle&\triangle&\circ&\circ&\triangle&\circ&\circ\\ \hline \triangle&\circ&\triangle&\triangle&\square&\circ&\triangle\\ \hline \circ&\circ&\triangle&\square&\square&\triangle&\triangle\\ \hline \end{array}$$
$$\begin{array}{|c|c|c|c|c|c|} \hline \square&\square&\circ&\triangle&\triangle&\circ&\circ\\ \hline \square&\circ&\circ&\triangle&\blacksquare&\blacksquare&\circ\\ \hline \triangle&\triangle&X&X&\blacksquare &\square &\square\\ \hline \triangle&\square&X&X&\triangle&\triangle&\square\\ \hline \square&\square&\circ&\circ&\triangle&\circ&\circ\\ \hline \circ&\circ&\triangle&\circ&\square&\circ&\triangle\\ \hline \circ&\triangle&\triangle&\square&\square&\triangle&\triangle\\ \hline \end{array}$$