[proofplan] We evaluate the binomial expansion of $(x+y)^n$ at the specific values $x=1$ and $y=1$. The [binomial theorem](/theorems/750) expresses this power as a sum whose coefficients are exactly the binomial coefficients in the $n$th row. Substituting $1$ for both variables turns the left-hand side into $2^n$ and the right-hand side into the required sum. [/proofplan] [step:Apply the binomial theorem to the $n$th power] Fix $n \in \mathbb{N} \cup \{0\}$. By the binomial theorem (citing a result not yet in the wiki: Binomial theorem), for all [real numbers](/page/Real%20Numbers) $x,y \in \mathbb{R}$, \begin{align*} (x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k. \end{align*} [guided] Fix an arbitrary $n \in \mathbb{N} \cup \{0\}$. Since $n$ is a nonnegative integer, the binomial theorem applies to the $n$th power of a sum. It states that for every pair of real numbers $x,y \in \mathbb{R}$, \begin{align*} (x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k. \end{align*} This identity is useful here because the coefficients appearing in the expansion are precisely the entries of the $n$th binomial row, namely $\binom{n}{0}, \binom{n}{1}, \dots, \binom{n}{n}$. [/guided] [/step] [step:Evaluate the expansion at $x=1$ and $y=1$] Set $x=1$ and $y=1$ in the binomial expansion. Since $1^{n-k}=1$ and $1^k=1$ for every integer $k$ with $0 \le k \le n$, we obtain \begin{align*} (1+1)^n = \sum_{k=0}^{n} \binom{n}{k}. \end{align*} The left-hand side is $2^n$, so \begin{align*} 2^n = \sum_{k=0}^{n} \binom{n}{k}. \end{align*} Reordering the equality gives \begin{align*} \sum_{k=0}^{n} \binom{n}{k} = 2^n. \end{align*} Because $n \in \mathbb{N} \cup \{0\}$ was arbitrary, the identity holds for every nonnegative integer $n$. [/step]