Question about Q-analogs I am trying to prove the following
Given $n \in \mathbb{N}$ we define $[n]_{q} = (1-q^{n})/(1-q)$.
We also define $[n]_{q} ! = [n-1]_{q} ! \cdot [n]_{q}$, with $[1]_{q} ! =1$.
Then I want to prove the following
Given compositions $\lambda$, $\mu$ we say $\mu \geq \lambda$ if
$$
\lambda_{1} = \mu_{1} + \ldots + \mu_{i_{1}}
$$
$$
\lambda_{2} = \mu_{i_{1}+1} + \mu_{i_{1}+2} + \ldots + \mu_{i_{2}}
$$
Etc, in other words if one sees $\lambda$ as a division of the interval $[1, \ldots , | \lambda | ]$ then $ \mu \geq \lambda$ if $\mu$ has the same divisions than $\lambda$ and possibly more.
For example $\lambda = (3,2,1)$ can be seen as $123|45|6$ and $\mu=2121$ can be seen as $12|3|45|6$. In this case $\mu \geq \lambda$.
Given a composition $\lambda$, we define
$$
f_{\lambda} (q) = \prod_{i=1}^{ \ell (\lambda) } [ \lambda_{i} ]_{q} !^{-1}
$$
I want to prove that
$$
f_{\lambda} \left( \dfrac{1}{q} \right) = \sum_{\mu \geq \lambda} f_{\mu} (q) (-1)^{\ell (\mu )- | \lambda |}
$$
As the function is defined multiplicatively, I know that this follows if I can prove it for $\lambda = (\lambda_{1})$.
Do anyone knows any reference in which I could look for theory that could aid me to come up with a proof?
 A: After trying again induction, but this time using all the previous terms, I think I have succeded.
We want to prove
\begin{equation}
\label{Eq1}
 f_{m} \left( \frac{1}{q} \right) = \sum_{ | \lambda | = m} f_{\lambda} (q) (-1)^{\ell ( \lambda ) + | \lambda |  } \qquad (1)
\end{equation}
First, we divide the compositions of $m$ on the following sets: the ones that end with a block of size $1$, they are $2^{m-1}$, the ones that end with a block of size $2$, they are $2^{m-2}$, and so. There is only $1$ that ends with a block of size $m-1$ and only one that ends with a block of size $m$.
So
$$
\sum_{ | \lambda | = m} f_{\lambda} (q) (-1)^{\ell ( \lambda ) + | \lambda | } = \sum_{0 \leq n < m}  \sum_{ | \mu | = n}  f_{\mu} (q) (-1)^{\ell ( \mu ) +1 + | \mu | +(m-n) } \dfrac{1}{ [m - n ]_{q} !}
$$
If we assume by strong induction that $(1)$ holds for $n<m$, then this is is equal to
$$
\sum_{0 \leq n < m}   (-1)^{m-n+1} f_{n} \left( \frac{1}{q} \right) \dfrac{1}{ [m - n ]_{q} !} = (-1)^{1+m} \sum_{0 \leq n < m}   (-1)^{n} f_{n} \left( \frac{1}{q} \right) \dfrac{1}{ [m - n ]_{q} !}
$$
We add $f_{m}(1/q)$ to this, and we obtain
$$
(-1)^{1+m} \sum_{0 \leq n < m}   (-1)^{n} f_{n} \left( \frac{1}{q} \right) \dfrac{1}{ [m - n ]_{q} !} +(-1)^{2m+1} f_{m}  \left( 
\frac{1}{q} \right) =
$$
$$
=
 (-1)^{1+m} \sum_{0 \leq n \leq m}   (-1)^{n} f_{n} \left( \frac{1}{q} \right) \dfrac{1}{ [m - n ]_{q} !}
$$
A straightfoward computation shows that $f_{n} (1/q) = q^{n(n-1)/2}$.
$$
(-1)^{1+m} \sum_{0 \leq n \leq m}   (-1)^{n} q^{n(n-1)/2} f_{n} \left( q \right) \dfrac{1}{ [m - n ]_{q} !}
$$
By definition $f_{n}(q) = [n]_{q}!^{-1}$, we replace on the expresion and obtain.
$$
(-1)^{1+m} \sum_{0 \leq n \leq m}   (-1)^{n} q^{(n-1)n/2} \dfrac{1}{ [n]_{q}! [m - n ]_{q} !}
$$
If we multiply this by $f_{m}(1/q)^{-1}=[m]_{q}! (-1)^{1+m}$, then we get
$$
 \sum_{0 \leq n \leq m}   (-1)^{n} q^{(n-1)n/2} \dfrac{ [m]_{q}! }{ [n]_{q}! [m - n ]_{q} !}
$$
Which by Gauss Binomial Formula (found on page 15, equation (5.5) of Quantum Calculus Victor Kac) with $a=-1$ and $x=1$ this is $0$.
So $ f_{m}(1/q) = \sum_{| \mu | = m} f_{\mu} (q) (-1)^{\ell ( \mu ) + | \lambda |}$.
