Prove $\frac{(q^n-1)(q^{n-1}-1)\dots (q^{n-r+1}-1)}{(q^r-1)(q^{r-1}-1)\dots(q-1)}=\sum_{\lambda\subseteq\Pi}q^{^{|\Pi\backslash\lambda|}}$ Prove the combinatorial identity
$$\frac{(q^n-1)(q^{n-1}-1)\dots (q^{n-r+1}-1)}{(q^r-1)(q^{r-1}-1)\dots(q-1)}=\sum_{\lambda\subseteq\Pi}q^{^{|\Pi\backslash\lambda|}},$$
where the summation is performed over all different Young diagrams $\lambda$ that fit into a rectangle $\Pi$ of size $r\times(n-r)$, and the exponent $|\Pi\backslash\lambda|$ is equal to the number of cells in the complement of the diagram to a rectangle (the empty diagram $\lambda=\emptyset$ and the entire rectangle $\lambda=\Pi$ are also taken into account).
The left side reminds me of formulas for the number of subspaces of a given dimension in an $n$-dimensional space over a field of $q$ elements. However, I am not sure if this is how one should reason.
 A: Let
$$
F(n,r)=\frac
{(q^n-1)(q^{n-1}-1)\cdots(q^{n-r+1}-1)}
{(q^r-1)(q^{r-1}-1)\cdots(q-1)}
,\qquad 
G(n,r)=\sum_{\lambda\subseteq\Pi_{r,n-r}}q^{|\Pi-\lambda|}$$
First of all, note that
$$
\begin{align}
F(n,r)&=\frac
{(q^n\color{blue}{-q^r+q^r}-1)(q^{n-1}-1)\cdots(q^{n-r+1}-1)}
{(q^r-1)(q^{r-1}-1)\cdots(q-1)}
\\&=\frac
{(q^{n}-\color{blue}{q^r})(q^{n-1}-1)\cdots(q^{n-r+1}-1)}
{(q^r-1)(q^{r-1}-1)\cdots(q-1)}
+\frac
{(\color{blue}{q^r}-1)(q^{n-1}-1)\cdots(q^{n-r+1}-1)}
{(q^r-1)(q^{r-1}-1)\cdots(q-1)}
\\&=q^r\cdot\frac{(q^{n-1}-1)\cdots(q^{n-r+1}-1)(q^{n-r}-\color{blue}{1})}{(q^r-1)(q^{r-1}-1)\cdots(q-1)}+\frac{(q^{n-1}-1)\cdots(q^{n-r+1}-1)}{(q^{r-1}-1)\cdots(q-1)},\end{align}
$$
We have shown
$$
F(n,r)= q^r F(n-1,r)+F(n-1,r-1)
$$
This makes it clear that $F(n,k)$ is actually a polynomial for all $n\ge k\ge 0$.
The key point is this; we can prove that $G(n,k)$ satisfies the same recurrence relation as $F(n,k)$. To prove this, let
$$
S_1=\{\lambda \subseteq \Pi\mid \text{width of $ \lambda$  is less than $n-r$}\}\\
S_2=\{\lambda \subseteq \Pi\mid \text{width of $ \lambda$  equals $n-r$}\}\hspace{.8cm}
$$
We can split $G(n,k)$ into $\sum_{S_1} q^{|\Pi-\lambda|}$ and $\sum_{S_2} q^{|\Pi-\lambda|}$.

*

*For $\sum_{S_1} q^{|\Pi-\lambda|}$, all of the summands $q^{\Pi-\lambda}$ have a factor of $q^r$, coming from the right column of the box that $\lambda$ does not touch. If you factor out that $q^r$, what remains is the sum of $q^{|\Sigma-\mu|}$, where $\Sigma$ is n $r\times (n-r-1)$ box and $\mu$ ranges over all partitions fitting inside $\Sigma$. Therefore,
$$
\sum_{S_1} q^{|\Pi-\lambda|}= q^r G(n-1,r)
$$


*Similarly, if you delete the bottom row from each $\lambda$ in $\sum_{S_2} q^{|\Pi-\lambda|}$, then what remains is a sum over partitions fitting in an $(r-1)\times (n-r)$ box of the size of the complement of that partitions, which is exactly $G(n-1,r-1)$.
We have proven the polynomial equation
$$
G(n,r)=q^rG(n-1,r-1)+G(n-1,r).
$$
After verifying that $G(n,k)$ and $F(n,k)$ satisfy the same base cases (both are equal to $1$ when $n=k=0$, and $0$ when $n=0$ but $k\neq 0$), the fact that they satisfy the same recurrence leads to a proof by induction on $n$ that $F(n,k)=G(n,k)$ for all $n,k\ge 0$.
In particular, since everything was done in the ring of integer polynomials over $q$, there is no need to appeal to the equation being true for infinitely many primes $q$. Once we have proven they are equal as polynomials, they are automatically equal as functions of $q$.
