Solution:

A 10-staircase has 10 "upper right corners" $P$, each of which must be the upper right corner of some rectangle, and no two of which can belong to the same rectangle. It also has a single lower left corner $Q$ which must belong to the same rectangle as one of the ten points $P$. Since this rectangle has one side of length 1, it must be a $10 \times 1$ rectangle placed either vertically or horizontally along the long side of the staircase. The remainder of the figure is then a 9-staircase to be filled with 9 rectangles.

We can then repeat the argument to find that one of the long sides of the 9-staircase must be filled by a $9 \times 1$ rectangle, leaving an 8-staircase. This continues until we reach the 1-staircase, a single square, which can be filled in only one way.

The placement of the $10 \times 1$ rectangle is irrelevant because of the symmetry of the shape. But the $9 \times 1$ through $2 \times 1$ rectangles each involve a choice between two alternatives, so the number of tilings is $2^{8} = 256$.