The Binomial Theorem is easier and more naturally proven in a combinatorics context but can be proven by induction.

Problem: Prove the Binomial Theorem by Induction.

Solution: Let $P(n)$ be the proposition that for $x,y\in\mathbb{R}$, $n\in\mathbb{N}$

$(x+y)^n=\sum_{i=0}^n{n\choose i}x^{n-i}y^i$

(Binomial Theorem)

$P(1)$:

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

($P(1)$ is true)

Now assume $P(k)$ is true; that is:

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

Now

$(x+y)^{k+1}=(x+y)^k(x+y)$

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

$=\underbrace{\left(\sum_{i=0}^k{k\choose i}x^{k+1-i}y^i\right)}_{=S_1}+\underbrace{\left(\sum_{i=0}^k{k\choose i}x^{k-i}y^{i+1}\right)}_{=S_2}$

Now all terms are of the form $c(i)x^{k+1-i}y^i$ as $i$ runs from $0\rightarrow k+1$. Let $j\in\{0,1,\dots,k+1\}$. Now the $x^{k+1-j}y^j$ term has constant from $S_1$ and $S_2$:

$x^{k+1-j}y^j\left[{k\choose j}+{k\choose j-1}\right]$

$\Rightarrow (x+y)^{k+1}=\sum_{i=0}^{k+1}\left[{k\choose j}+{k\choose j-1}\right]x^{k+1-j}y^j$

It is a straightforward exercise to show:

${n+1\choose k}={n\choose k}+{n\choose k-1}$

Hence

$(x+y)^{k+1}=S_1+S_2=\sum_{i=0}^{k+1}{k+1\choose i}x^{k+1-i}y^ii$

($P(k+1)$ is true)

$P(1)$ is true. $P(k)\Rightarrow P(k+1)$. Hence $P(n)$ is true for all $n\in\mathbb{N}$; i.e. the Binomial Theorem is true $\bullet$