First we prove that $\cidl{a}_i + \prod_{j \neq i} \cidl{a}_j = R$ for each $i$ Without loss of generality, assume that $i = 1$ Then $$ R = (\cidl{a}_1 + \cidl{a}_2) (\cidl{a}_1 + \cidl{a}_3) \dotsm (\cidl{a}_1 + \cidl{a}_n) , $$ since each factor $\cidl{a}_1 + \cidl{a}_j$ is $R$ Expanding the product, each term will contain $\cidl{a}_1$ as a factor, except the term $\cidl{a}_2 \cidl{a}_2 \dotsm \cidl{a}_n$ So we have $$ (\cidl{a}_1 + \cidl{a}_2) (\cidl{a}_1 + \cidl{a}_3) \dotsm (\cidl{a}_1 + \cidl{a}_n) \subseteq \cidl{a}_1 + \cidl{a}_2 \cidl{a}_2 \dotsm \cidl{a}_n , $$ and hence the expression on the right hand side must equal $R$

Now we can prove that $\prod \cidl{a}_i = \bigcap \cidl{a}_i$ by induction. The statement is trivial for $n = 1$ For $n = 2$ note that $$ \cidl{a}_1 \cap \cidl{a}_2 = (\cidl{a}_1 \cap \cidl{a}_2) R = (\cidl{a}_1 \cap \cidl{a}_2) (\cidl{a}_1 + \cidl{a}_2) \subseteq \cidl{a}_2 \cidl{a}_1 + \cidl{a}_1 \cidl{a}_2 = \cidl{a}_1 \cidl{a}_2 , $$ and the reverse inclusion is obvious, since each $\cidl{a}_i$ is an ideal. Assume that the statement is proved for $n-1$ and condsider it for $n$ Then $$ \bigcap_1^n \cidl{a}_i = \cidl{a}_1 \cap \bigcap_2^n \cidl{a}_i = \cidl{a}_1 \cap \prod_2^n \cidl{a}_i , $$ using the induction hypothesis in the last step. But using the fact proved above and the $n = 2$ case, we see that $$ \cidl{a}_1 \cap \prod_2^n \cidl{a}_i = \cidl{a}_1 \cdot \prod_2^n \cidl{a}_i = \prod_1^n \cidl{a}_i . $$

Finally, we are ready to prove the Chinese remainder theorem. Consider the ring homomorphism $R \to \prod R/\cidl{a}_i$ defined by projection on each component of the product: $x \mapsto (\cidl{a}_1 + x, \cidl{a}_2 + x, \dots , \cidl{a}_n + x)$ It is easy to see that the kernel of this map is $\bigcap \cidl{a}_i$ which is also $\prod \cidl{a}_i$ by the earlier part of the proof. So it only remains to show that the map is surjective.

Accordingly, take an arbitrary element $(\cidl{a}_1 + x_1, \cidl{a}_2 + x_2, \dots, \cidl{a}_n + x_n)$ of $\prod R/\cidl{a}_i$ Using the first part of the proof, for each $i$ we can find elements $y_i \in \cidl{a}_i$ and $z_i \in \prod_{j \neq i} \cidl{a}_j$ such that $y_i + z_i = 1$ Put $$ x = x_1 z_1 + x_2 z_2 + \dots + x_n z_n. $$ Then for each $i$ $$\cidl{a}_i + x = \cidl{a}_i + x_i z_i,$$ since $x_j z_j \in \cidl{a}_i$ for all $j \neq i$ $$=\cidl{a}_i + x_i y_i + x_i z_i,$$ since $x_i y_i \in \cidl{a}_i$ $$=\cidl{a}_i + x_i (y_i + z_i) = \cidl{a}_i + x_i \cdot 1 = \cidl{a}_i + x_i.$$ Thus the map is surjective as required, and induces the isomorphism $$\frac{R}{\prod \cidl{a}_i} \to \prod \frac{R}{\cidl{a}_i}.$$