Binomial Theorem

Numbers
Let $$x, y \in \mathbb{R}$$.

Then $$\forall n \in \mathbb{Z}_+: \left({x+y}\right)^n = \sum_{k=0}^n {n\choose k}x^{n-k}y^k$$.

Ring Theory
Let $$\left({R, +, \odot}\right)$$ be a ringoid such that $$\left({R, \odot}\right)$$ is a commutative semigroup.

Let $$k, n \in \mathbb{Z}: n \ge 2$$. Then:

$$\forall x, y \in R: \odot^n \left({x + y}\right) = \odot^n x + \sum_{k=1}^{n-1} \binom n k \left({\odot^{n-k} x}\right) \odot \left({\odot^k y}\right) + \odot^n y$$

where $$\binom n k = \frac {n!} {k! \left({n-k}\right)!}$$ (see Binomial Coefficient).

If $$\left({R, \odot}\right)$$ has an identity element, then:

$$\forall x, y \in R: \odot^n \left({x + y}\right) = \sum_{k=0}^n \binom n k \left({\odot^{n-k} x}\right) \odot \left({\odot^k y}\right)$$

Proof
Base Case:

$$n=0$$

$$(x+y)^0=1={0\choose 0}x^{0-0}y^0=\sum_{k=0}^0 {0\choose k}x^{0-k}y^k$$

Therefore the base case holds.

Inductive Hypothesis:

$$(x+y)^j = \sum_{k=0}^j {j\choose k}x^{j-k}y^k$$ for all $$j \ge 1$$

Inductive Step:

And so we are done by the Principle of Mathematical Induction.

The proof for the Abstract Algebra version follows the same strategy.