[proofplan] We prove the two endpoint identities by evaluating the defining Bernstein sum at $x=0$ and $x=1$. At $x=0$, every term with $k \geq 1$ contains the vanishing factor $0^k$, leaving only the $k=0$ term. At $x=1$, every term with $k \leq n-1$ contains the vanishing factor $(1-1)^{n-k}$, leaving only the $k=n$ term. [/proofplan]

[step:Evaluate the Bernstein polynomial at the left endpoint]Since $n \in \mathbb N$, and Androma's convention is $\mathbb N=\{1,2,3,\dots\}$, we have $n \geq 1$. By definition of $B_n f$, substituting $x=0$ gives \begin{align*} B_n f(0)=\sum_{k=0}^{n} \binom{n}{k} f\left(\frac{k}{n}\right)0^k(1-0)^{n-k}. \end{align*} For every integer $k$ with $1 \leq k \leq n$, the factor $0^k$ is $0$, so those terms vanish. The remaining term is the term with $k=0$. In polynomial evaluation the zero exponent factor is interpreted as $0^0=1$, and since $n \geq 1$ the node satisfies $0/n=0$. Therefore the remaining term equals \begin{align*} \binom{n}{0} f\left(\frac{0}{n}\right)0^0(1-0)^n = f(0). \end{align*} Hence \begin{align*} B_n f(0)=f(0). \end{align*}[/step]

[guided]We start by recording the convention on $n$. Since $n \in \mathbb N$, and Androma's convention is $\mathbb N=\{1,2,3,\dots\}$, we have $n \geq 1$. In particular, division by $n$ is valid and $0/n=0$. Now substitute the endpoint $x=0$ into the definition of the Bernstein polynomial. This gives \begin{align*} B_n f(0)=\sum_{k=0}^{n} \binom{n}{k} f\left(\frac{k}{n}\right)0^k(1-0)^{n-k}. \end{align*} The point of evaluating at $0$ is that the factor $x^k$ separates the $k=0$ term from all later terms. If $1 \leq k \leq n$, then $0^k=0$, so each summand with $k \geq 1$ is equal to $0$. Thus only the $k=0$ summand remains. For $k=0$, we use the polynomial convention that the zero exponent factor is $0^0=1$. We also have $\binom{n}{0}=1$, $0/n=0$ because $n \geq 1$, and $(1-0)^n=1$. Hence the remaining term is \begin{align*} \binom{n}{0} f\left(\frac{0}{n}\right)0^0(1-0)^n = f(0). \end{align*} Therefore \begin{align*} B_n f(0)=f(0). \end{align*}[/guided]

[step:Evaluate the Bernstein polynomial at the right endpoint] By definition of $B_n f$, substituting $x=1$ gives \begin{align*} B_n f(1)=\sum_{k=0}^{n} \binom{n}{k} f\left(\frac{k}{n}\right)1^k(1-1)^{n-k}. \end{align*} For every integer $k$ with $0 \leq k \leq n-1$, the exponent $n-k$ is positive, so the factor $(1-1)^{n-k}$ is $0$. Hence all those terms vanish. The remaining term is the term with $k=n$. Its factor $(1-1)^{n-n}$ is the zero exponent factor $0^0$, interpreted as $1$ in polynomial evaluation. Therefore it equals \begin{align*} \binom{n}{n} f\left(\frac{n}{n}\right)1^n(1-1)^0 = f(1). \end{align*} Thus \begin{align*} B_n f(1)=f(1). \end{align*} Combining the two endpoint evaluations proves the theorem. [/step]