It suffices to show that either $P$ or $Q$ is non-singular, and I'll do it for $P$. By subtracting multiples of the first row from all other rows, we get to the matrix
$$
\begin{pmatrix} 2 & 1 & 1 & 1 & 1 & \ldots & 1& 1 \\
-3 & 2 & 0 & 0 & 0 & \ldots & 0 & 0\\
-5 & -1 & 3 & 0 & 0 & \ldots & 0 & 0 \\ -7 & -2 & -1 & 4 & 0 & \ldots & 0 & 0 \\
&&&\ldots &&&\\ -(2n-1) & -(n-2) & -(n-3) & -(n-4) & -(n-5) & \ldots & -1 & n
\end{pmatrix}
$$
In fact, the precise values don't matter: The key feature is that all entries in the lower triangular part are negative, and those on the diagonal and in the first row are positive.

This means that if we now subtract $1/n$ times the last row from the first row, then the remaining entries in the first row stay positive. So we now have to *subtract* the $(n-1)$st row to get rid of the $(1,n-1)$ element, and again, this only makes the remaining entries in the first row bigger.

We continue in this way, always subtracting rows from the first row. When we're done with this, we are still looking at a positive $(1,1)$ element. In particular, it is not zero, and the transformed matrix has full rank.