Let $A$ be a commutative complete ring with unit for the $I$-adic topology, where $I$ is the ideal of $A$. Let $(M_n)_{n\geq 0}$ be $A$-modules such that $I^{n+1}M_n=0$ and that there exist a surjective homomorphism $\pi_n:M_{n+1}\to M_n$ with $\ker (\pi_n)=I^{n+1}M_{n+1}$. Let $M=\varprojlim_nM_n$ and denote the surjective canonical homomorphism by $u_n:M\to M_n$. Fix $d\geq 0$. How can I show that for any $n\geq d$, there is an exact sequence $$0\to I^{d+1}M_n\to M_n\to M_d\to 0$$
Qing Liu: Algebraic Geometry and Arithmetic Curves ex 1.3.11 a
One can prove the claim $$M_n/I^{d+1} M_n = M_d$$ by descending induction on $d$ with fixed $n$. First $$0 \to I^{n} M_{n} \to M_{n} \to M_{n-1} \to 0$$ proves $M_{n}/I^{n}M_n = M_{n-1}$ so the claim is true for $d = n - 1$.
Next by induction $$M_n/I^{d+1} M_n = M_d$$ and therefore $$(I^d M_n)/(I^{d+1}M_n) = I^d M_d$$ Now $$ \begin{multline} M_n/I^d M_n = (M_n/I^{d+1} M_n)/(I^d M_n/I^{d+1} M_n) = M_d/I^d M_d = M_{d-1} \end{multline} $$ P.S. Indeed it was not necessary to bring the completion $M$ and even the completeness of $A$ into play. Also $I^{n+1}M_n = 0$ was not used.