Is NP-hard closed against cartesian product with arbitrary languages?

If $L_1$ is NP-hard, $L_1 \times L_2$ is NP-hard for every $L_2 \neq \emptyset$, where

$\qquad \displaystyle L_1 \times L_2 = \{(w_1,w_2) \mid w_1 \in L_1, w_2 \in L_2\}$

Is it true or false and why?

I can't prove it but I also don't find counter example.

$L_1 \times L_2$ is NP-hard.

Proof: Let $L$ be a language in NP, let $f$ be a reduction of $L$ to $L_1$ and let $w_2$ be in $L_2$. Define $g(x) = (f(x), w_2)$. Now $g$ is a polynomial time many-to-one reduction of $L$ to $L_1 \times L_2$ because clearly:

$x \in L \quad \Longleftrightarrow\quad (f(x), w_2) \in L_1\times L_2$

StackExchange Computer Science Q#2489, answer score: 6