Recall the four following rules:

• odd plus odd and even plus even is even

• even plus odd is odd

• even times anything is even

• odd times odd is odd

These rules can be easily verified by representing arbitrary odd numbers as $2m+1$ and arbitrary even numbers as $2n$ respectively, for integers $m,n.$

With this in mind, let's examine each answer choice individually:

A:

Note that $3$ is odd. Modulo $2$, this gives us $$1 + 1 \cdot 1 \equiv 0 \pmod 2.$$ From our above rules, we know that this is even.

B: $$1 \cdot 1 - 1 \equiv 0 \pmod 2.$$ Once again, this is even.

C: $$1 \cdot 1^2 + 1 \cdot 1^2 \equiv 0 \pmod 2.$$ This is also even.

D: $$(1 \cdot 1 + 1)^2 \equiv 0 \pmod 2.$$ Unfortunately, this is also even.

E: $$1 \cdot 1 \cdot 1 \equiv 1 \pmod 2.$$ This is odd.

Therefore, E is the only answer choice that is an odd integer.

Thus, E is the correct answer.