Chinese remainder theorem for pairwise comaximal ideals

First take $r=2$. Choose $u\in I_1$ and $v\in I_2$ with $u+v=1$. If $x\in I_1\cap I_2$, then $x=xu+xv\in I_1{I}_2$, so $I_1\cap I_2=I_1{I}_2$. For classes $a+I_1$ and $b+I_2$, the element $av+bu$ satisfies $av+bu\equiv a(\mathrm{m}\mathrm{o}\mathrm{d}{I}_1)$ and $av+bu\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}{I}_2)$, so the canonical map $R\to R\mathrm{/}{I}_1\times R\mathrm{/}{I}_2$ is surjective with kernel $I_1\cap I_2=I_1{I}_2$.

Now assume $r\ge 2$. Fix $i$ and put $J_i={\prod }_{j\mathrm{\ne }i}{I}_j$. For each $j\mathrm{\ne }i$, choose $u_j\in I_i$ and $v_j\in I_j$ with $u_j+v_j=1$. Expanding ${\prod }_{j\mathrm{\ne }i}(u_j+v_j)=1$ shows that $1-{\prod }_{j\mathrm{\ne }i}{v}_j\in I_i$ and ${\prod }_{j\mathrm{\ne }i}{v}_j\in J_i$, so $I_i+J_i=R$. Choose $e_i\in J_i$ with $e_i\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}{I}_i)$. Then $e_i\in I_j$ for every $j\mathrm{\ne }i$. Given residue classes $a_i+I_i$, the element $x:={\sum }_{i=1}^r{a}_i{e}_i$ satisfies $x\equiv a_i(\mathrm{m}\mathrm{o}\mathrm{d}{I}_i)$ for every $i$. Hence the canonical map $R\to {\prod }_{i=1}^{r}R\mathrm{/}{I}_i$ is surjective.

The kernel of the canonical map is plainly ${\bigcap }_{i=1}^r{I}_i$. To compare this with the product, induct on $r$. The case $r=1$ is tautological and the case $r=2$ is step 1.1. For $r>2$, let $K={\bigcap }_{i=2}^r{I}_i$. By the induction hypothesis, $K={\prod }_{i=2}^r{I}_i$. Step 2.1 with $i=1$ gives $I_1+K=R$, so applying the two-ideal case to $I_1$ and $K$ yields ${\bigcap }_{i=1}^r{I}_i=I_1\cap K=I_{1}K={\prod }_{i=1}^r{I}_i$. Therefore $R\mathrm{/}{\prod }_{i=1}^r{I}_i\cong {\prod }_{i=1}^{r}R\mathrm{/}{I}_i$.