We understand both $\prod_{n = 1}^N I_{E_n}$ and $I_{\bigcap_{n = 1}^N E_n}$ as functions from $X$ to $\{ 0, 1 \}$. Since the domain and codomain sets are the same, it suffices to show that they attain the same values on $X$.

If $x \in X$, then one has the following chain of equivalencies
\begin{equation}
\begin{split}
\left( \prod_{n = 1}^N I_{E_n} \right)(x) = I_{E_1}(x) \cdots I_{E_N}(x) = 1 \quad & \iff \quad x \in E_1, \quad \dots, \quad x \in E_N \\
& \iff \quad x \in \bigcap_{n = 1}^N E_n \\
& \iff \quad I_{\bigcap_{n = 1}^N}(x) = 1 \\
\end{split}
\end{equation}
The claim then follows from

R2965: Indicator function is uniquely identified by its support. $\square$