Show that the regular language is contained in another Let L1 and L2 be two non empty regular language. How can I show that the following expression is correct with an example ? 
$\bigotimes$ denote the transducer cross product.

$L_1^+ \bigotimes L_2^+$ contains $(L_1 \bigotimes L_2)^+$

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Thanks PHPNick. Would my version (rewritten from yours) be correct ? I am just trying to make it more explicit to verify my doubts. If my understanding of the cross product and the + operator is correct.

Let $L_1$ and $L_2$ be the following

$L_1 = \{u_1,u_2,u_3\} $ $L_2 = \{v_1,v_2,v_3\} $

Then $L_1^+$ and $L_2^+$ would be the set of all strings that can be made by concatenating finitely many (and at least one) strings (corrected by David Richerby)

$L_1^+ = \{u_1,u_1u_1,u_1u_1u_1,...,u_2,...,u_3,..., u_1u_2,u_1u_2u_1,\dots\} $

$L_2^+ = \{v_1,v_1v_1,v_1v_1v_1,...,v_2,...,v_3,..., v_1v_2,v_1v_2v_1,\dots\} $

The two sets defined in the question are thus

$L_1^+ \bigotimes L_2^+ = \{ (u_1,v_1), (u_1,v_1v_1), \dots, (u_1u_1,v_1), \dots, (u_1u_1,v_1v_1), \dots \}$
$(L_1 \bigotimes L_2)^+ = \{ (u_1,v_1), (u_1,v_2), (u_1,v_3), \dots, (u_2,v_1), \dots \}$

The set $L_1^+ \bigotimes L_2^+$ thus contains $(L_1 \bigotimes L_2)^+$ because we can see that $(u_1,v_1), (u_1,v_2)$ etc are present in $L_1^+ \bigotimes L_2^+$
It is just a simple subset proof:

Let $(u, v) \in (L_1 \otimes L_2)^+$. Thus, there is a decomposition of $(u, v)$ into $(u_1 \ldots u_k, v_1 \ldots v_k)$ with $u_i \in L_1$ and $v_i \in L_2$ and $k &gt; 0$. Hence, we have $u \in L_1^+, v \in L_2^+$. So overall we have $(u, v) \in L_1^+ \otimes L_2^+$.