Consider the modified sequence defined by $t_i=r_i-1$. All of its elements are integers which are at least $-1$ and whose total sum is $-k$, and you want to prove there are exactly $k$ rotations whose partial sums are all greater than $-k$.

We induct on $n$, the claim being obvious when $n=0$. If $n\ge 1$, then some entry in the list is nonnegative (because there are more than $k$ terms suming to $-k$). Let $a$ be a nonnegative entry of the list, and rotate the list to begin with $a$. Call the rotated list $u_1,u_2,\dots,u_{n+k}$, with $u_1=a$.

Let $s_i=u_1+u_2+\dots+u_i$. Note that

This means that $s_i$ must attain every value between $a$ and $-k$. In particular, there is a partial sum equal to $0$. Let $l$ be the smallest index for which $s_l=0$, so in the sublist $(u_1,u_2,\dots,u_l)$, all partial sums are positive except for the whole sum which is negative.

I claim that when counting rotations of the original list $(t_1,\dots,t_{n+k})$ whose partial sums are all at least $-k$, none of the entries $(u_1,\dots,u_l)$ can occur at the end. If $u_j$ occurred at the end, we would have this situation:
$$
(\underbrace{u_{j+1},u_{j+2},\dots,u_l}_{\text{sum is $\le 0$}},
\;\underbrace{u_{l+1},u_{l+2},\dots,u_{n+k}}_{\text{sum is $-k$}},
\;\underbrace{u_{1},u_{2},\dots,u_{j}}_{\text{sum is $\ge 0$}},)
$$
but then the partial sum $u_{j+1},u_{j+2},\dots,u_{n+k}$ would be less than or equal to $-k$.

Now consider what happens when you delete $(u_1,\dots,u_l)$ from the list. What remains is a list of fewer integers which still sum to $-k$. By induction, exactly $k$ rotations of this list have all partial sums greater than $-k$. Furthermore, re-inserting $(u_1,\dots,u_l)$ to the immediate left of $u_{l+1}$ will preserve this property (why?). Therefore, exactly $k$ rotations of the original list have all partial sums greater than $-k$.

## Best Answer

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 $q$.