Here's an elementary proof that doesn't require ultrafilters, but uses axiom of choice.

The group 

$\prod_{n=1}^\infty \mathbb{Z} / \bigoplus_{n=1}^\infty \mathbb{Z}$ 

clearly surjects onto

$\prod_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z}) / \bigoplus_{n=1}^\infty (\mathbb{Z}/p\mathbb{Z})$.


The latter is a nontrivial $\mathbb{F}_p$-vector space, being a $p$-torsion abelian group. Therefore, we can choose a basis $\{e_i\}_{i\in I}$. Finally, we get a map into $\mathbb{Z}/p\mathbb{Z}$ by killing all but one of the $e_i$.