I have a tower of abelian groups where for a certain $A_n$, the long exact sequence here holds:
$0 \to \underset{\longleftarrow}{\lim}_n A_n \longrightarrow \underset{n}{\prod} A_n \stackrel{\partial}{\longrightarrow} \underset{n}{\prod} A_n \longrightarrow \underset{\longleftarrow}{\lim}^1_n A_n \to 0$
I also know that for my particular $A_n$, first term in the sequence (the inverse limit) is trivial. Thus, $\partial$ is injective. Does that mean that $\partial$ is an isomorphism? This would be based on the assumption that both products are the same size. Is this false?
No, the inverse limit being trivial doesn't tell you anything about $\lim^1$. For instance, consider the system with $A_n=\mathbb{Z}$ for all $n$ and all the maps are multiplication by $2$; then the inverse limit is trivial since the only element of $\mathbb{Z}$ that is infinitely divisible by $2$ is $0$. Then $\partial$ will be the map $\mathbb{Z}^\mathbb{N}\to\mathbb{Z}^\mathbb{N}$ that sends $(a_0,a_1,a_2,\dots)$ to $(a_0-2a_1,a_1-2a_2,a_2-2a_3,\dots)$. This is not surjective: for instance, its image does not contain the sequence $(1,0,1,0,\dots)$ that alternates between $1$s and $0$s. (Proving this is a nice exercise.)