What are the group homomorphisms from $ \prod_ {n \in \mathbb {N}} \mathbb {Z} / \bigoplus_ {n \in \mathbb {N}} \mathbb {Z} $ to $ \mathbb {Z} $? By a theorem of Specker, there’s only the zero map since any map out of $ \prod_{n \in \mathbb{N}} \mathbb{Z} $ is determined by the values of the unit vectors, which all lie in $ \bigoplus_{n \in \mathbb{N}} \mathbb{Z} $, but the original proof is more general, uses a bunch of machinery, and in German. Isn’t there an easier way?

There is a nice quick proof. I'm not sure who the proof is due to.
The statement is equivalent to

If $P$ is the group of sequences ${\bf a}=(a_0,a_1,\dots)$ of integers, and $f:P\to\mathbb{Z}$ is a homomorphism that vanishes on finite sequences (so that $f({\bf a})=f({\bf b})$ whenever ${\bf a}$ and ${\bf b}$ differ in only finitely many places), then $f=0$.

Suppose $f:P\to\mathbb{Z}$ is a homomorphism that vanishes on finite sequences. 
For any ${\bf a}\in P$, we can write $a_n=b_n+c_n$, where $b_n$ is divisible by $2^n$ and $c_n$ is divisible by $3^n$.
Then for each $n$, ${\bf b}$ differs in only finitely many places from a sequence divisible by $2^n$, so $f({\bf b})$ is divisible by $2^n$ for all $n$, and so $f({\bf b})=0$. Similarly $f({\bf c})=0$, and so $f({\bf a})=f({\bf b}+{\bf c})=0$.
