Proving B* = B on a given set I have the set:

B = {x ∈ {0,1}* | there is an equal number of 0's and 1's in x}

and therefore,

B* = {e,01,10,0011,0101,0110,1100,1010,1001,....etc}

I need to either prove or disprove that B*=B

I believe they are equal, because B* is just the concatenation of one string onto another. I think the trick is since the 0's and 1's have to be of equal number, that concatenating strings would keep the same number of 0's and 1's the same. 

I just need a little help in the right direction as to prove B*=B. I was thinking of showing that if 

B⊆B* and B*⊆B, that B*=B holds.

Any help would be much appreciated!
(Criterion of no-op Kleene star) Let $V$ be a language such that $\epsilon\in V$. Then $V=V^*$ if and only if $V=VV$, where $V^*$ is the Kleene star.

Proof: "$\Longrightarrow$":  $V=V\epsilon\subseteq VV$ while $VV\subseteq V^*=V$.

"$\Longleftarrow$". $V\subseteq V^*$ by definition. Let $V_i$ as defined in Wikipedia. Since $V_0\subseteq V$ and assuming $V_n\subseteq V$, $V_{n+1}=V_nV\subseteq VV=V$, we know $V_i\subseteq V$ for all $i\ge0$ by induction. Hence $$V^*=\bigcup_{i=0}^\infty V_{i+1}\subseteq \bigcup_{i=0}^\infty V=V\,.$$



Exercise 1. Show that $\epsilon\in B$ and $B=BB$.

Exercise 2. Let $V$ be a non-empty language such that $V=VV$. Show that $\epsilon\in V$. (Hence for a non-empty language $V$, $V=V^*$ if and only if $V=VV$.)

Exercise 3. Show that $V^*=(V^*)^*$.