Prove that a finite Boolean ring is isomorphic to $\mathbb Z_2\times \mathbb Z_2\times\cdots\times \mathbb Z_2$ [duplicate]

If $R$ is a finite Boolean ring with $1\neq 0$ then prove that $R\cong \mathbb Z_2\times \mathbb Z_2\times \cdots\times \mathbb Z_2$.

How should I proceed? Please give some hints only.

Hints:
1. $R$ is abelian.
2. $R$ is a unitary $\mathbb{Z}/2\mathbb{Z}$-module.
3. $R$ is a unitary free $\mathbb{Z}/2\mathbb{Z}$-module.

Let $R$ be a boolean ring. Clearly, $R$ is abelian and has characteristic $2$. Therefore, $R$ is a unitary $\mathbb{Z}/2\mathbb{Z}$-module, or equivalently, a $\mathbb{Z}/2\mathbb{Z}$-vector space, because $\mathbb{Z}/2\mathbb{Z}$ is a field. Since every vector space over a field $F$ is a unitary free $F$-module, $R$ is a unitary free $\mathbb{Z}/2\mathbb{Z}$-module. Hence, $R$ is a direct sum of some copies of $\mathbb{Z}/2\mathbb{Z}$. If $R$ is, in addition, finite, then $R\cong\left(\mathbb{Z}/2\mathbb{Z}\right)^{\oplus n}=\left(\mathbb{Z}/2\mathbb{Z}\right)^n$ as a $\mathbb{Z}/2\mathbb{Z}$-vector space for some $n\in\mathbb{N}$. We shall induct on $n$ that $R\cong \left(\mathbb{Z}/2\mathbb{Z}\right)^{\oplus n}$ as a ring. If $n=0$ or $n=1$, then the claim is trivial. If $n>1$, then there exists an element $e\in R\setminus\{0,1\}$. Clearly, $R=Re\oplus R(1-e)$, where $Re$ and $R(1-e)$ are boolean subrings of $R$, which are also ideals of $R$, with strictly smaller dimensions (over $\mathbb{Z}/2\mathbb{Z}$). The induction hypothesis is applied to $Re$ and $R(1-e)$, so $Re\cong\left(\mathbb{Z}/2\mathbb{Z}\right)^{\dim_{\mathbb{Z}/2\mathbb{Z}}\big(Re\big)}$ and $R(1-e)\cong\left(\mathbb{Z}/2\mathbb{Z}\right)^{\dim_{\mathbb{Z}/2\mathbb{Z}}\big(R(1-e)\big)}$ as rings. Consequently, we get $R\cong \left(\mathbb{Z}/2\mathbb{Z}\right)^{\dim_{\mathbb{Z}/2\mathbb{Z}}\big(Re\big)+\dim_{\mathbb{Z}/2\mathbb{Z}}\big(R(1-e)\big)}=\left(\mathbb{Z}/2\mathbb{Z}\right)^{\dim_{\mathbb{Z}/2\mathbb{Z}}(R)}=\left(\mathbb{Z}/2\mathbb{Z}\right)^n$ as a ring.

Hint: {0, 1} comprises a subring (with unit) of $R$ with two elements. A ring with unit with just 2 elements is a field. Observe that $R$ is a finite-dimensional vector space over this field.