Consider the functions
$ f(x) = \prod_{n=1}^{\infty}(1-x^n) $
and
$ g(x) = \prod_{n=1}^{\infty}(1+x^n) $
$f(x)$ is defined for $x\in[-1,1]$ and $g(x)$ is defined for $x\in[-1,0]$.
I was wondering if there was a different/better way to express these functions, such as a closed-form expression/power series.
Desmos screenshot ($f(x)$ in red and $g(x)$ in blue)
Edit: As pointed out by donaastor the Pentagonal Number Theorem can be used to express $f(x)$ as a power series.
Edit 2: The Euler function $\phi(x)$ seems similar to $f(x)$ but with a different domain. For $\phi(x)$ the domain is $x\in(-\infty,-1)\cup(1,\infty)$.
Note that for $x\in(-1,1)$ we have $-1<x^n<1$, so $0<1-x^n<2$. Thus $$f(x)=\left|\prod_{n\ge1}(1-x^n)\right|=\prod_{n\ge1}|1-x^n|=\prod_{n\ge1}(1-x^n).$$ Furthermore, $$f(x)g(x)=\prod_{n\ge1}(1+x^n)(1-x^n)=\prod_{n\ge1}(1-x^{2n})=f(x^2),$$ so $g(x)=f(x^2)/f(x)$.
These functions don't have closed forms in terms of elementary functions, but they are extensively studied for their number-theoretic properties. We will consider their series expansions.
It is the case that these functions have a lot of significance in the study of integer partitions. Specifically, they are related to generating functions of specific kinds of partitions. I'll demonstrate a general result and show how it relates to these functions.
Theorem: Let $T\subseteq\Bbb N$ be nonempty and let $p_T(n)$ be equal to the number of integer partitions of a positive integer $n$ into parts all belonging to $T$ (also known as $T$-partitions of $n$). Then the generating function of $p_T(n)$ is given by $$P_T(q):=\sum_{n\ge0}p_T(n)q^n=\prod_{k\in T}\frac{1}{1-q^k},\tag{$\star$}$$ where $|q|<1$.
Immediately you will see that when $T=\Bbb N$, $P_{\Bbb N}(q)=1/f(q)$, and $g(q)=P_{\Bbb N}(q)/P_{\Bbb N}(q^2)$. In this case, $P_{\Bbb N}(q)$ is the generating function for the total number of partitions of $n$. The proof of the result is at the end of this answer.
Since $P(x)=1/f(x)=\sum_{n\ge0}p(n)x^n,$ we have that if $f(x)=\sum_{n\ge0}f_nx^n$ then $$P(x)f(x)=\frac1{f(x)}f(x)=1=\sum_{n\ge0}x^n\sum_{k=0}^{n}p(k)f_{n-k}.$$ This gives $$p(0)=f_0=1,\qquad\qquad \sum_{k=0}^np(k)f_{n-k}=0\qquad n\ge1.$$ As others noted, this is the pentagonal number theorem, and $f_n=(-1)^k$ when $n=k(3k-1)/2$ and $0$ otherwise.
As for $g(q)=f(q^2)/f(q)=P(q)/P(q^2)$, we will examine $P(q)$ again. Note that $$\begin{align} P(q)=\prod_{n\ge1}\frac{1}{1-q^n}&=\frac{1}{1-q}\frac{1}{1-q^2}\frac{1}{1-q^3}\frac{1}{1-q^4}\cdots\\ &=\left(\frac{1}{1-q^2}\frac{1}{1-q^4}\frac{1}{1-q^6}\cdots\right)\left(\frac{1}{1-q}\frac{1}{1-q^3}\frac{1}{1-q^5}\cdots\right)\\ &=\left(\prod_{k\ge1}\frac{1}{1-q^{2k}}\right)\left(\prod_{k\ge1}\frac{1}{1-q^{2k-1}}\right)\\ &=P(q^2)\prod_{k\ge1}\frac{1}{1-q^{2k-1}}. \end{align}$$ Thus $$g(q)=\frac{P(q)}{P(q^2)}=\prod_{k\ge1}\frac{1}{1-q^{2k-1}}.$$ Note however that if we write $O=\{2k-1:k\in\Bbb N\}$ to be the set of odd positive integers, we get that $$g(q)=\prod_{k\in O}\frac{1}{1-q^k}.$$ Applying the above theorem, $g(q)$ is seen to be the generating function for partitions of $n$ into odd parts.
Proof. Consider the series expansion of the product in $(\star)$. From the geometric series, we obtain $$\prod_{k\in T}\frac{1}{1-q^k}=\prod_{k\in T}\left(\sum_{n\ge0}q^{nk}\right).$$ Since $T\subseteq\Bbb N$, we have $k\ge1$, and thus the sums on the right converge absolutely, and we are justified in expanding the product as $$\prod_{k\in T}\left(\sum_{n\ge0}q^{nk}\right)=\sum_{n_1,n_2,n_3,...} q^{n_1k_1}q^{n_2k_2}q^{n_3k_3}\cdots=\sum_{n_{k_1},n_{k_2},n_{k_3},...}\prod_{k\in T}q^{n_kk},\tag1$$ where the sum is over all $n_{k_1},n_{k_2},n_{k_3},...\ge0$, and we have written $T=\{k_1,k_2,k_3,...\}$.
Suppose that $T$ is finite. Then write $T=\{k_1,...,k_m\}$ to obtain $$\prod_{k\in T}\frac{1}{1-q^k}=\sum_{n_1,...,n_m\ge0}q^{n_1k_1+n_2k_2+...+n_mk_m}.\tag2$$ We will use this later.
In the case that $T$ is infinite, we consider the limit $$\prod_{k\in T}q^{n_kk}=\lim_{i\to\infty}q^{n_1k_1}q^{n_2k_2}\cdots q^{n_ik_i}=\lim_{i\to\infty}q^{n_1k_1+n_2k_2+...+n_ik_i}.$$ To evaluate this limit we must consider the behavior of the sequence $(n_i)_{i\ge1}=(n_1,n_2,n_3,...)$. If it converges to zero, then there is some $j\ge1$ such that $i> j\Rightarrow n_i=0$. In this case $$i>j\Rightarrow n_1k_1+n_2k_2+...+n_jk_j+...=n_1k_1+n_2k_2+...+n_jk_j,$$ which gives $$\prod_{k\in T}q^{n_kk}=\lim_{i\to\infty}q^{n_1k_1+n_2k_2+...+n_jk_j}=q^{n_1k_1+n_2k_2+...+n_jk_j}.$$ In the case that $(n_i)_{i\ge1}$ does not converge to $0$, the sum $n_1k_1+n_2k_2+...+n_ik_i$ grows without bound. Since $|q|<1$, we have $$\prod_{k\in T}q^{n_kk}=\lim_{i\to\infty}q^{n_1k_1+n_2k_2+...+n_ik_i}=\lim_{i\to\infty}q^i=0.$$ So we only need to consider the cases of $(n_i)_{i\ge1}$ with a finite number of nonzero terms. That is, $$\prod_{k\in T}\frac{1}{1-q^k}=\sum_{m\ge0}\sum_{n_1,...,n_m\ge0}q^{n_1k_1+n_2k_2+...+n_mk_m}.\tag3$$ So, regardless of the size of $T$, we are taking a sum of $q^N$, where $N\ge0$ is an integer and it can be written as $N=n_1k_1+n_2k_2+...+n_mk_m$ for some choice of $(n_i,k_i)$. Clearly, this is an integer partition of $N$. Examining this partition, we see $$n_1k_1+n_2k_2+..+n_mk_m=\underbrace{k_1+k_1+...+k_1}_{n_1\text{ times}}+...+\underbrace{k_m+k_m+...+k_m}_{n_m\text{ times}},$$ revealing that this is a partition of $N$ into parts belonging to $T$, as $k_i\in T$. We may reindex the sums $(2),(3)$ in terms of N, and get that they are both equal to $$\prod_{k\in T}\frac{1}{1-q^k}=\sum_{N\ge0}\sum_{N=n_1k_1+...+n_rk_r}q^N=\sum_{N\ge0}q^N\sum_{N=n_1k_1+...+n_rk_r}1.$$ The quantity $\sum_{N=n_1k_1+...+n_rk_r}1$ clearly counts the number of partitions of $N$ into parts belonging to $T$. That is, $$\prod_{k\in T}\frac{1}{1-q^k}=\sum_{n\ge0}p_T(n)q^n=P_T(q).$$ This completes the proof.