I'm trying exercise $A.39$ in Fulton and Harris. They suggest to first prove the formula
$|x_j^{l_i}| \prod_{j=1}^k(1-x_j)^{-1} = \sum |x_j^{m_i}| \hspace{1in} (\ast)$
where the sum on the right is over all $k$ - tuples $(m_1,\ldots,m_k)$ of non-negative integers with $m_1 \geq l_1\geq \ldots \geq m_k \geq l_k$. We induct on $k$:
Now the base case $k=1$ for $(\ast)$ is easy to see. So to see how my induction argument would work in general (using the inductive hypothesis to finish the problem) I looked at how the $k=3$ case can be reduced to the $k=2$ case. I then applied the "inductive hypothesis" to the $k = 2$ case. If I understand this, I am done. For $k = 3$, we have:
$\begin{eqnarray*} \frac{1}{(1-x_1)(1-x_2)(1-x_3)}\left|\begin{array}{ccc} x_1^{l_1} & x_2^{l_1} & x_3^{l_1} \\ x_1^{l_2}& x_2^{l_2} & x_3^{l_2} \\ x_1^{l_3} & x_2^{l_3} & x_3^{l_3} \end{array}\right| &=& \frac{x_1^{l_1} \left|\begin{array}{cc} x_2^{l_2} & x_3^{l_2} \\ x_2^{l_3} & x_3^{l_3} \end{array}\right| - x_2^{l_1}\left|\begin{array}{cc} x_1^{l_2} & x_3^{l_2} \\ x_1^{l_3} & x_3^{l_3} \end{array}\right| + x_3^{l_1}\left|\begin{array}{cc} x_1^{l_2} & x_2^{l_2} \\ x_1^{l_3} & x_2^{l_3} \end{array}\right| }{(1-x_1)(1-x_2)(1-x_3)} \\ &=& \frac{x_1^{l_1}}{(1-x_1)} \sum_{(m_2,m_3):m_2 \geq l_2 \geq m_3 \geq l_3} \left|\begin{array}{cc} x_2^{m_2} & x_3^{m_2} \\ x_2^{m_3} & x_3^{m_3} \end{array}\right| \\ &&- \frac{x_2^{l_1}}{(1-x_2)}\sum_{(m_2,m_3):m_2 \geq l_2 \geq m_3 \geq l_3} \left|\begin{array}{cc} x_1^{m_2} & x_3^{m_2} \\ x_1^{m_3} & x_3^{m_3} \end{array}\right| \\ &&+ \frac{x_3^{l_1}}{(1-x_3)}\sum_{(m_2,m_3):m_2 \geq l_2 \geq m_3 \geq l_3} \left|\begin{array}{cc} x_1^{m_2} & x_2^{m_2} \\ x_1^{m_3} & x_2^{m_3} \end{array}\right| \end{eqnarray*}$
where the last step was made by applying the "inductive hypothesis" to $k =2$ case. I can see that I am nearly done, however:
Problem: The three sums are taken for $(m_2,m_3) : m_2 \geq l_2 \geq m_3 \geq l_3$. This means that $m_2$ can possibly be unbounded. However to finish the problem I need that the sums be taken over those $m_2$ constrained to be less than $l_1$. How can I get around this?
Thanks.