Tiling of a deficient $7\times7$ chessboard with L trominoes 
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.
 A: Consider this (i am going draw some formal pictures after I have taken a nap.)
$$\begin{array}{|c|c|c|c|c|c|}
\hline
X&X&\circ&\circ&\triangle&\triangle&\circ\\
\hline
X&X&\circ&\square&\triangle&\circ&\circ\\
\hline
\circ&\circ&\square&\square&\blacksquare &\blacksquare &\square\\
\hline
\circ&\triangle&\circ&\circ&\blacksquare&\square&\square\\
\hline
\triangle&\triangle&\circ&\square&\square&\triangle&\triangle\\
\hline
\circ&\circ&\triangle&\square&\circ&\triangle&\square\\
\hline
\circ&\triangle&\triangle&\circ&\circ&\square&\square\\
\hline
\end{array}$$
$$\begin{array}{|c|c|c|c|c|c|}
\hline
\square&\square&\circ&\triangle&\triangle&\circ&\circ\\
\hline
\square&\circ&\circ&\triangle&\blacksquare&\blacksquare&\circ\\
\hline
X&X&\square&\square&\blacksquare &\square &\square\\
\hline
X&X&\circ&\square&\triangle&\triangle&\square\\
\hline
\triangle&\triangle&\circ&\circ&\triangle&\circ&\circ\\
\hline
\triangle&\circ&\triangle&\triangle&\square&\circ&\triangle\\
\hline
\circ&\circ&\triangle&\square&\square&\triangle&\triangle\\
\hline
\end{array}$$
$$\begin{array}{|c|c|c|c|c|c|}
\hline
\square&\square&\circ&\triangle&\triangle&\circ&\circ\\
\hline
\square&\circ&\circ&\triangle&\blacksquare&\blacksquare&\circ\\
\hline
\triangle&\triangle&X&X&\blacksquare &\square &\square\\
\hline
\triangle&\square&X&X&\triangle&\triangle&\square\\
\hline
\square&\square&\circ&\circ&\triangle&\circ&\circ\\
\hline
\circ&\circ&\triangle&\circ&\square&\circ&\triangle\\
\hline
\circ&\triangle&\triangle&\square&\square&\triangle&\triangle\\
\hline
\end{array}$$
A: You need only check the yellow cases, as all other cases can be found by a trivial symmetry:

