Graded algebra and Hilbert–Poincaré series, closed formula, need some more detail.

Question

In a now deleted question on MathOverflow here, I asked the following question.

Let $A$ be a $k$-algebra equipped with a grading $A = \oplus_{i \in \mathbb{Z}}A_i$, as a $k$-vector space.

Question. Where can I find a reference to derivations of closed formulas for the formal power series$$P(A, t) := \sum_{i = 0}^\infty (\dim_k A_i) \cdot t^i$$in the cases where:

1. $A = k[x_1, \ldots, x_n]$,
2. $A = k\langle x_1, \ldots, x_n\rangle$,

and $n \ge 1$? Or would anybody be so kind as to supply a quick derivation themselves?

Qiaochu Yuan gave the following answer in the comments.

These are straightforward combinatorics exercises.

1. Poincaré series are multiplicative under tensor product, so the problem reduces to the case $n = 1$, where the Poincaré series is ${1\over{1 - t}}$. So in general it's ${1\over{(1 - t)^n}}$.
2. The number of words on $n$ letters of length $k$ is $n^k$, so the answer is ${1\over{1 - nt}}$.

The question was put on hold, presumably because it was at too low of a level for MathOverflow. However, I don't quite understand Qiaochu's answer, is it possible someone could add a bit more detail as to how he got his answers?

Answer 1

First, you should be sure you understand that $P(A\otimes B,t)=P(A,t)P(B,t)$. Now, $$K[x_1,\ldots,x_n]=K[x_1]\otimes K[x_2]\otimes\cdots\otimes K[x_n]$$ so the first problem reduces to computing $P(K[x],t)$. Well, the $k$th degree component of $K[x]$ is one-dimensional, spanned by $x^k$, so $$P(K[x],t)=1+t+t^2+t^3+\cdots=\frac{1}{1-t}.$$ It follows that $P(K[x_1,\ldots,x_n],t)=\frac{1}{(1-t)^n}$.

As for the free algebra $K\langle x_1,\ldots,x_n\rangle$, as Qiaochu noted, the number of words of length $k$ is $n^k$. As these words are linearly independent, the dimension of the $k$th degree component is $n^k$. So, $$P(K\langle x_1,\ldots,x_n\rangle,t)=1+nt+n^2t^2+n^3t^3+\cdots=\frac{1}{1-nt}.$$