Infinite primes proof There is a proof for infinite prime numbers that i don't understand.

right in the middle of the proof:
"since every such $m$ can be written in a unique way as a product of the form:
$\prod_{p\leqslant x}p^{k_p}$. we see that the last sum is equal to: $\prod_{\binom{p\leqslant x}{p\in \mathbb{P}}}(\sum_{k\leqslant 0}\frac{1}{p^k})$.
I don't see that. can anyone can explain this step to me?

It's not every such $m$ that can be written as $\prod_{\substack{p\in\mathbb{P},\\ p\leq x}}\sum_{k\geq0}\frac{1}{p}$, it's that the sum of all such $m$ is the same as the product and summation. What your proof is saying are equal is $$\prod_{\substack{p\in\mathbb{P},\\p\leq x}}\sum_{k\geq0}\frac{1}{p}=\sum_{\substack{m \text{ with}\\ \text{prime}\\ \text{factors}\\\leq x}}\frac{1}{m}$$
For example's sake, say $x=6$. Then we're summing over $m$ with prime factors of $2$, $3$, $5$. Listing these out, we have something like
$$\prod_{p\in\{2,3,5\}}\sum_{k=0}^\infty \frac{1}{p^k}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{2^2}+\frac{1}{5}+\frac{1}{2\cdot 3}+\frac{1}{2^3}+\frac{1}{3^2}+\frac{1}{2\cdot 5}++\frac{1}{2^2\cdot3}+\dots\text{ }.$$
