Kleene closure, concatenation problem If $L_1 = \emptyset$ , $L_2= \{a\}$
then what is $$L_1\cdot L_2^* \cup L_1^*$$

The answer given is $\{\epsilon\}$ but I think it should be $\{\epsilon,a\}$.
My Approach : 
$L_1^* = \{\epsilon\}$ 
$L_2^* = \{\epsilon,a\}$ 
$L_1\cdot L_2^* = \{\epsilon,a\}$ 
$L_1\cdot L_2^* \cup L_1^* = \{\epsilon,a\}$.

Where's my mistake?
For any language $L$, $$\emptyset⋅L=\emptyset$$.

Therefore, 
$$L_1\cdot L_2^* = \emptyset \cdot L_2^* = \emptyset$$

And thus $$L_1\cdot L_2^*\cup L_1^* = \emptyset \cup L_1^*= L_1^* = \{\epsilon\}$$.