Poincaré series of an $A$-module I'm studying Atiyah and Macdonald's "Introduction to Commutative Algebra" and I'm having some problems computing the Poincaré series of an $A$-module.
Even the Example after $11.3$, which is dealt very quickly on the text, does not appear clear to me: 

Why can we conclude that $P(A,t)=(1-t)^{-s}$ (where $A=A_0[x_1,\ldots,x_s]$ with $A_0$ Artin ring and $x_1,\ldots,x_s$ independent indeterminates) ?

Do you know any source of guided excercises and\or examples which can help me to get more confident with these objects?
 A: Atiyah-Macdonald explain that $l(A_n)=\binom{s+n-1}{s-1}$ (by writing down an explicit basis consisting of monomials). It follows
$$P(A,t) = \sum_{n=0}^{\infty} \binom{s+n-1}{s-1} t^n.$$
We claim that this equals $(1-t)^{-s}$ (as formal power series), and prove this by induction on $s$. The case $s=1$ is the usual geometric series. For the induction step, we start with
$$(1-t)^{-s} = \sum_{n=0}^{\infty} \binom{s+n-1}{s-1} t^n$$
and apply $\frac{d}{dt}$ on both sides. It follows
$$s (1-t)^{-(s+1)} = \sum_{n=0}^{\infty} \binom{s+n}{s-1} (n+1) t^n$$
The claim for $s+1$ follows from
$$\binom{s+n}{s} = \frac{n+1}{s} \cdot \binom{s+n}{s-1}.$$
We can also give a more direct proof as follows: Taking the $s$th power in $(1-t)^{-1} = \sum_{n=0}^{\infty} t^n$, we get
$$(1-t)^{-s} = \sum_{n=0}^{\infty} \left(\sum_{k_1+\dotsc+k_s = n} 1 \cdot \dotsc \cdot 1 \right) t^n = \sum_{n=0}^{\infty} \binom{s+n-1}{s-1} t^n.$$
A: Since this question concerns a typo in Atiyah-MacDonald which has not been pointed out in the answers nor the comments, I add a new answer for future reference. The Poincaré series in the example should be
$$ P(A, t) = \text{len}(A_0) (1-t)^{-s}.$$
This follows from the fact, explained in the example, that $A_n$ is free over $A_0$ of rank $\binom{n+s-1}{s-1}$ and length is additive (in particular $\text{len}(A_0^k) = k \text{len} (A_0)$). Hence, $\text{len}(A_n) = \text{len}(A_0) \binom{n+s-1}{s-1}$ and $$P(A,t) =\text{len}(A_0) \sum_{n \geq 0} \binom{n+s-1}{s-1}t^n =\text{len}(A_0) (1-t)^{-s}.$$
