, 2020

Solution:

Viewing the problem in reverse, it is equivalent to show that we can use multiple $n$-staircases to make a single, larger $n$-staircase, because that larger $n$-staircase is made up of strictly smaller $n$-staircases, and is the example we need.

For the construction, we first attach two $n$-staircases of the same size together to make an $n \times (n+1)$ rectangle. Then, we arrange $n(n+1)$ of these rectangles in a $(n+1) \times n$ grid, giving an $n(n+1) \times n(n+1)$ size square. Finally, we can use $\frac{n(n+1)}{2}$ of these squares to create a larger $n$-staircase of $n^{2}(n+1)^{2}$ smaller staircases, so we are done.

Solution 2:

An alternative construction using only $2n+2$ staircases was submitted by team Yeah Knights A. We provide a diagram for $n=5$ and allow the interested reader to fill in the details.