[proofplan] Expand $(a+b)^n$ as a product of $n$ identical factors and encode each summand by the subset of factors from which the term $b$ is selected. Because multiplication in $\mathbb{R}$ is commutative, all selections with exactly $k$ copies of $b$ give the same monomial $a^{n-k}b^k$. Counting the subsets of $\{1,\dots,n\}$ of size $k$ gives the coefficient $\binom{n}{k}$, and summing over $k=0,\dots,n$ gives the stated identity. [/proofplan] [step:Encode every term in the expanded product by a subset of factors] Fix $a,b \in \mathbb{R}$ and $n \in \mathbb{N}$. Let $I_n := \{1,\dots,n\}$. Since $n$ is finite, repeated use of the distributive law in $\mathbb{R}$ gives one summand for each choice of either $a$ or $b$ from each factor in \begin{align*} (a+b)^n=\prod_{i=1}^{n}(a+b). \end{align*} For each subset $S \subset I_n$, let $S$ record precisely the indices of the factors from which $b$ is chosen. The corresponding [real number](/pages/1303) is \begin{align*} T_S := \left(\prod_{i \in I_n \setminus S} a\right)\left(\prod_{i \in S} b\right), \end{align*} with the convention that an empty product is $1$. Therefore \begin{align*} (a+b)^n=\sum_{S \subset I_n} T_S. \end{align*} [guided] We fix $a,b \in \mathbb{R}$ and $n \in \mathbb{N}$, because the theorem asserts the identity for arbitrary such choices. Define the finite index [set](/pages/1142) $I_n := \{1,\dots,n\}$, whose elements label the $n$ factors in the product \begin{align*} (a+b)^n=\prod_{i=1}^{n}(a+b). \end{align*} When this finite product is expanded, the distributive law says that each summand is obtained by making one choice from each factor: either choose $a$ from the $i$-th factor or choose $b$ from the $i$-th factor. A subset $S \subset I_n$ records exactly this data by declaring that $i \in S$ means “choose $b$ from the $i$-th factor,” while $i \in I_n \setminus S$ means “choose $a$ from the $i$-th factor.” For such a subset $S$, define the corresponding real number \begin{align*} T_S := \left(\prod_{i \in I_n \setminus S} a\right)\left(\prod_{i \in S} b\right), \end{align*} where an empty product is interpreted as $1$. This convention covers the two endpoint cases: $S=\varnothing$, where every factor contributes $a$, and $S=I_n$, where every factor contributes $b$. Since every expansion choice gives exactly one subset $S \subset I_n$, and every subset $S \subset I_n$ gives exactly one expansion choice, the full expansion is \begin{align*} (a+b)^n=\sum_{S \subset I_n} T_S. \end{align*} [/guided] [/step] [step:Group the expansion by the number of chosen copies of $b$] For each integer $k$ with $0 \le k \le n$, define \begin{align*} \mathcal{S}_k := \{S \subset I_n : |S|=k\}. \end{align*} If $S \in \mathcal{S}_k$, then $|I_n \setminus S|=n-k$. Since multiplication in $\mathbb{R}$ is commutative, \begin{align*} T_S = \left(\prod_{i \in I_n \setminus S} a\right)\left(\prod_{i \in S} b\right) = a^{n-k}b^k. \end{align*} The subsets of $I_n$ are partitioned by their cardinalities, so \begin{align*} (a+b)^n = \sum_{S \subset I_n} T_S = \sum_{k=0}^{n}\sum_{S \in \mathcal{S}_k} T_S = \sum_{k=0}^{n} |\mathcal{S}_k|\,a^{n-k}b^k. \end{align*} [guided] The expansion from the previous step is indexed by all subsets $S \subset I_n$. To identify the coefficient of a fixed monomial, we group these subsets according to how many copies of $b$ they choose. For each integer $k$ satisfying $0 \le k \le n$, define \begin{align*} \mathcal{S}_k := \{S \subset I_n : |S|=k\}. \end{align*} If $S \in \mathcal{S}_k$, then exactly $k$ factors contribute $b$. Since $I_n$ has $n$ elements, the complement $I_n \setminus S$ has $n-k$ elements, so exactly $n-k$ factors contribute $a$. Because $a$ and $b$ are [real numbers](/page/Real%20Numbers) and multiplication in $\mathbb{R}$ is commutative, the order of these chosen factors does not affect the product. Hence \begin{align*} T_S = \left(\prod_{i \in I_n \setminus S} a\right)\left(\prod_{i \in S} b\right) = a^{n-k}b^k. \end{align*} Every subset $S \subset I_n$ has exactly one cardinality $k \in \{0,\dots,n\}$, so the families $\mathcal{S}_0,\dots,\mathcal{S}_n$ partition the set of all subsets of $I_n$. Therefore the expansion can be regrouped as \begin{align*} (a+b)^n = \sum_{S \subset I_n} T_S = \sum_{k=0}^{n}\sum_{S \in \mathcal{S}_k} T_S = \sum_{k=0}^{n} |\mathcal{S}_k|\,a^{n-k}b^k. \end{align*} [/guided] [/step] [step:Count the subsets of each size and substitute the binomial coefficient] For each integer $k$ with $0 \le k \le n$, the number of subsets of $I_n$ with cardinality $k$ is \begin{align*} |\mathcal{S}_k|=\frac{n!}{k!(n-k)!}=\binom{n}{k}. \end{align*} Indeed, ordered lists of $k$ distinct elements of $I_n$ can be counted in two ways: directly there are \begin{align*} n(n-1)\cdots(n-k+1)=\frac{n!}{(n-k)!} \end{align*} such lists, while each $k$-element subset gives exactly $k!$ orderings. Thus \begin{align*} |\mathcal{S}_k|\,k! = \frac{n!}{(n-k)!}, \end{align*} and division by $k!$ gives the formula above. Substituting this count into the grouped expansion yields \begin{align*} (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k. \end{align*} This is the claimed identity. [/step]