Show that if $F_{1}(x_{1}) \cdots F_{n}(x_{n}) = F(x_{1}, \ldots, x_{n})$, then $X_{1}, \ldots, X_{n}$ are independent

Let $X_{1}, \ldots, X_{n}$ be random variables. FOr each $k \in \{1, \ldots, n\}$, the cumulative distribution function for $X_{k}$ is the function $F_{k}$ defined on $\mathbb{R}$ by $F_{k}(x) = P(X_{k} \leq x)$. The joint cumulative distribution function for $X_{1}, \ldots, X_{n}$ is the function $F$ defind on $\mathbb{R}^n$ by $$F(x_{1}, \ldots, x_{n}) = P(X_{1} \leq x_{1}, \ldots, X_{n} \leq x_{n}).$$We want to prove that if for all $x_{1}, \ldots, x_{n} \in \mathbb{R}$, we have $F(x_{1}, \ldots, x_{n}) = F_{1}(x_{1}) \cdots F_{n}(x_{n})$, then $X_{1}, \ldots, X_{n}$ are independent.
For each $k = 1, \ldots, n$, we define the $\pi$-system $$\mathscr{E}_{k} = \{(-\infty,x_{k}]: x_{k} \in \mathbb{R}\}.$$It's clearly closed under intersection. So it's indeed a $\pi$ -system. By assumption, we have $$F(x_{1}, \ldots, x_{n}) = F(x_{1}) \cdots F(x_{n}).$$For all $(-\infty,x_{1}] \in \mathscr{E}_{1}, \ldots, (-\infty,x_{n}] \in \mathscr{E}_{n}$, we have $$P(X_{1} \leq x_{1}, \ldots, X_{n} \leq x_{n}) = P(X_{1} \leq x_{1}) \cdots P(X_{k} \leq x_{k}).$$Thus, the $\pi$-system $\mathscr{E}_{1}, \ldots, \mathscr{E}_{n}$ are independent, which implies that $\sigma(\mathscr{E}_{1}), \ldots, \mathscr{E_{k}}$ are also independent. We then want to show $\sigma(\mathscr{E}_{k}) = \sigma(X_{k})$. By definition, $\sigma(X_{k}) = \{X_{k}^{-1}(B):B \in \mathcal{B}(\mathbb{R})\}$. We first show that $\sigma(\mathscr{E}_{k}) \subseteq \sigma(X_{k})$. For each $(-\infty,x_{k}] \in \mathscr{E}_{k}$, we have $X_{k}^{-1}((-\infty,x_{k}]) = \{\omega \in \Omega: X_{k}(\omega) \leq x_{k}\}$. I'm not sure how to go from here to show $\sigma(\mathscr{E_{k}}) \subseteq \sigma(X_{k})$. Please help! Thanks!