[proofplan] We use the monomial definition of homogeneity. A [homogeneous polynomial](/page/Homogeneous%20Polynomial) of degree $d$ is a finite $k$-linear combination of monomials whose exponent vectors have total degree $d$. Since $d>0$, every such exponent vector has at least one positive component, so every appearing monomial has a factor that vanishes at the origin. Therefore the whole finite sum evaluates to $0$. [/proofplan] [step:Write the homogeneous polynomial as a finite sum of degree $d$ monomials] Let $\mathbb{N}_0 := \mathbb{N} \cup \{0\}$. Since $f$ is homogeneous of degree $d$, there exist a finite set $A \subset \mathbb{N}_0^n$ and coefficients $c_a \in k$, indexed by $a=(a_1,\ldots,a_n) \in A$, such that \begin{align*} f=\sum_{a \in A} c_a x_1^{a_1}\cdots x_n^{a_n} \end{align*} and \begin{align*} a_1+\cdots+a_n=d \end{align*} for every $a \in A$. [/step] [step:Evaluate each appearing monomial at the origin] Fix $a=(a_1,\ldots,a_n) \in A$. Since $d \in \mathbb{N}$, we have $d>0$. From \begin{align*} a_1+\cdots+a_n=d \end{align*} and $a_i \in \mathbb{N}_0$ for each $i$, there exists an index $j \in \{1,\ldots,n\}$ such that $a_j>0$. Hence the evaluation of the corresponding monomial at the origin contains the factor $0^{a_j}=0$, and therefore \begin{align*} 0^{a_1}\cdots 0^{a_n}=0. \end{align*} [guided] Fix an exponent vector $a=(a_1,\ldots,a_n) \in A$. The total degree condition says \begin{align*} a_1+\cdots+a_n=d. \end{align*} Because $d \in \mathbb{N}$, the degree is positive. If every component $a_i$ were equal to $0$, then the sum $a_1+\cdots+a_n$ would be $0$, contradicting $d>0$. Therefore at least one index $j \in \{1,\ldots,n\}$ satisfies $a_j>0$. Now evaluate the monomial $x_1^{a_1}\cdots x_n^{a_n}$ at $(0,\ldots,0)$. The $j$-th factor becomes $0^{a_j}$, and this equals $0$ because $a_j>0$. A product in the field $k$ with one factor equal to $0$ is $0$, so \begin{align*} 0^{a_1}\cdots 0^{a_n}=0. \end{align*} [/guided] [/step] [step:Sum the monomial evaluations to obtain the value of $f$] Using the finite expansion of $f$ and evaluating at $(0,\ldots,0)$ gives \begin{align*} f(0,\ldots,0)=\sum_{a \in A} c_a 0^{a_1}\cdots 0^{a_n}. \end{align*} By the preceding step, every summand is equal to $0$. Hence \begin{align*} f(0,\ldots,0)=0. \end{align*} This is the desired conclusion. [/step]