How to prove $L \cdot L^{*} = L^{+}$
To summarize the comments.

A. Sometimes $L^+$ is defined to be $L\circ L^*$, where '$\circ$' is the  concatenations operator.

B. Assume the following definitions,


  $L^* \equiv \{\varepsilon\} \cup L \cup L^2 \cup \cdots$
  
  $L^+\equiv L \cup L^2 \cup \cdots$


Then, by the properties of the concatenation operator, $L\circ L^i = L^{i+1}$. Explicitly for the case of $i=0$, it also holds that $L\circ\{\epsilon\} = L$.

Then,
$$\begin{align}
L\circ L^* &amp;= L \circ \left(  \{\varepsilon\} \cup L \cup L^2 \cup \cdots \right )
\\ &amp;= (L\circ \{\varepsilon\}) \cup (L\circ L) \cup (L \circ L^2) \cup \cdots \\ &amp;= L \cup L^2 \cup \cdots \\ &amp;\equiv L^+
\end{align}$$

The only missing part is to explain the second transition--The distributivity of the concatenation operator over unions.