Proving $A^\ast = A$ on a given set
What you have written (i.e., $\{ \varepsilon, 01, 0011, 000111, \ldots \}$) is simply $A$ itself.

(Assuming $A^\ast$ is the Kleene star operation) you cannot prove $A = A^\ast$ because it is not correct.

The Kleene star $A^\ast$ is defined as the union of the powers $A^i$, where $A^0 = \{ \varepsilon \}$ and $A^{i+1} = A^i \cdot A = \{ w \cdot w' \mid w \in A^i, w' \in A \}$. An inductive argument easily shows:
$$A^i = \{ w_1 \cdots w_i \mid w_1, \ldots, w_i \in A \}
= \{ 0^{n_1} 1^{n_1} \cdots 0^{n_i} 1^{n_i} \mid n_1, \ldots, n_i \ge 0 \}$$

Since $A^\ast$ is the union over all such $A^i$, we obtain
$$A^\ast
= \{ 0^{n_1} 1^{n_1} \cdots 0^{n_i} 1^{n_i} \mid i, n_1, \ldots, n_i \ge 0 \},$$
which is obviously not equal to $A$.