Here is the problem:
Let $A$ be a Noetherian ring and $a$, $b$ be ideals in $A$. If $M$ is any $A$-module, let $M^a$, $M^b$ denote its $a$-adic and $b$-adic completions respectively. If $M$ is finitely generated, prove that $(M^a)^b\cong M^{a+b}$.
Since $A$ is Noetherian, by proposition 10.13, we only need to prove $(A^a)^b\cong A^{a+b}$. Now I don't know how to prove one of the hints.
How to prove: $$\varprojlim_m ( \varprojlim_n A/(a^nA+b^mA))= \varprojlim_n A/(a^nA+b^nA)$$
So this is my question.
Update: My guess is that: $$\varprojlim_m ( \varprojlim_n A/(a^nA+b^mA))= \varprojlim_{n,m} A/(a^nA+b^mA) =\varprojlim_n A/(a^nA+b^nA)$$
Sketch.
Facts. For a flat $A$-algebra $B$, let $I$ be an ideal of $A$, $M$ an $A$-module. Then $B\otimes_AIM\cong I(B\otimes_AM)$ (considering them as submodules of $B\otimes_AM$)
So when $B=\widehat{A}$ and $A$ is Noetherian, $M$ is finite, $\mathfrak{b}$ is ideal of $A$, we obtain $(\mathfrak{b}^mM)^{\mathfrak{a}}=\mathfrak{b}^mM^{\mathfrak{a}}$.
Take $\mathfrak{a}$-adic completions of the following exact sequence: $$ 0\to \mathfrak{b}^mM\to M\to M/\mathfrak{b}^mM\to 0, $$ we get a canonical isomorphism $$ (M/\mathfrak{b}^mM)^{\mathfrak{a}}\cong M^{\mathfrak{a}}/(\mathfrak{b}^mM)^{\mathfrak{a}}\cong M^{\mathfrak{a}}/\mathfrak{b}^mM^{\mathfrak{a}} $$
Thus we get the canonical isomorphism $$ (M^{\mathfrak{a}})^{\mathfrak{b}}\cong \varprojlim_m(M/\mathfrak{b}^mM)^{\mathfrak{a}}=\varprojlim_m\varprojlim_n M/(\mathfrak{b}^m+\mathfrak{a}^n)M=\varprojlim M/(\mathfrak{a}+\mathfrak{b})^kM. $$
Unfortunately, the map is NOT an isomorphism of topological groups in general.
Example. Consider the polynomial ring $k[x,y]$ in two variables, the ideal $(x,y)=(x)+(y)$, so the two rings $k[[x]][[y]]$ and $k[[x,y]]$ are isomorphic as rings, but one can check the isomorphism is not an isomorphism as topological rings.