Proof of the binomial theorem by induction

Proof

We argue by induction on $n$. For an integer $c$ and $r \in R$, $c \cdot r$ is the repeated sum. It satisfies $(c + d) \cdot r = c \cdot r + d \cdot r$ and $s (c \cdot r) = c \cdot (s r)$ for $s \in R$.

1. For $n = 0$, the left side is $(x + y)^0 = 1$. The right side is $\binom{0}{0} x^0 y^0 = 1$.
2. Suppose $(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k}$ for some $n$.
3. Then $(x + y)^{n+1} = (x + y)(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x \, x^k y^{n-k} + \sum_{k=0}^{n} \binom{n}{k} y \, x^k y^{n-k}$, by distributivity.
4. The ring is commutative, so $x \, x^k y^{n-k} = x^{k+1} y^{n-k}$ and $y \, x^k y^{n-k} = x^k y^{n+1-k}$.
5. Put $j = k + 1$ in the first sum. It becomes $\sum_{j=1}^{n+1} \binom{n}{j-1} x^j y^{n+1-j}$. The second sum is $\sum_{j=0}^{n} \binom{n}{j} x^j y^{n+1-j}$.
6. The first sum has the term $j = n + 1$ only once. It is $\binom{n}{n} x^{n+1}$, and $\binom{n}{n} = 1 = \binom{n+1}{n+1}$ by the factorial formula.
7. The second sum has the term $j = 0$ only once. It is $\binom{n}{0} y^{n+1}$, and $\binom{n}{0} = 1 = \binom{n+1}{0}$ by the definition of the binomial coefficient.
8. For $1 \le j \le n$, the two terms with $x^j y^{n+1-j}$ add to $\left( \binom{n}{j-1} + \binom{n}{j} \right) x^j y^{n+1-j}$.
9. By Pascal's rule with $k = j - 1$, this coefficient is $\binom{n+1}{j}$.
10. Collecting steps 6 to 9, $(x + y)^{n+1} = \sum_{j=0}^{n+1} \binom{n+1}{j} x^j y^{n+1-j}$. This completes the induction.