How to derive this infinite product formula?

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$$\prod_{n=0}^{\infty}\left(1 + x^{2^n}\right) = \frac{1}{1-x}$$

I tried numerous things, multiplying by $x$, dividing, but none of that worked. Also, I realized that:

$$\prod_{n=0}^{\infty} \left(1 + x^{2^n}\right) = \sum_{n=0}^{\infty} x^n$$

But I cannot prove the relation. I get:

$$(1 + x)(1+x^2)(1+x^4)(1+x^8)...$$

But for a general $n$ it is more difficult.

Since: $$\begin{eqnarray*} (1-x)(1+x)&=&(1-x^2),\\ (1-x^2)(1+x^2) &=& (1-x^4), \\ (1-x^4)(1+x^4) &=& (1-x^8),\\ \ldots\end{eqnarray*} $$ we have: $$ (1-x)\prod_{k=0}^{N}\left(1+x^{2^k}\right) = 1-x^{2^{N+1}} \tag{1}$$ so, assuming $|x|<1$ and letting $N\to +\infty$ we have: $$ (1-x)\prod_{k=0}^{+\infty}\left(1+x^{2^k}\right) = 1\tag{2} $$ as wanted.

Notice that a possible combinatorial interpretation of $(2)$ is: there is only one way of writing a natural number as a sum of distinct powers of two.

We have

$$\prod_{n = 0}^\infty (1 + x^{2^n}) = \lim_{m \to \infty} \prod_{n = 0}^m \frac{1 - x^{2^{n+1}}}{1 - x^{2^n}} = \lim_{m\to \infty} \frac{1 - x^{2^{m+1}}}{1 - x} = \frac{1}{1 - x}$$

for $|x| < 1$.

The partial products are $$1+x,$$ $$(1+x)+x^2(1+x)=1+x+x^2+x^3,$$ $$(1+x+x^2+x^3)+x^4(1+x+x^2+x^3)=1+x+x^2+x^3+x^4+x^4+x^5+x^6+x^7$$ $$\cdots$$ Every time the number of terms doubles. Doesn't that ring a bell ?