[proofplan] We use the coefficient description of $\mathcal{P}_d(k)$. First we show that this set is closed under addition and scalar multiplication, so it is a $k$-subspace of $k[x]$. Then we prove that the monomials $1,x,\dots,x^d$ span $\mathcal{P}_d(k)$ by unpacking the definition, and that they are linearly independent by uniqueness of coefficients in the [polynomial ring](/page/Polynomial%20Ring) $k[x]$. The dimension formula follows from the definition of dimension as the cardinality of a finite basis. [/proofplan]

[step:Show that bounded-degree polynomials are closed under vector space operations]Let $p,q \in \mathcal{P}_d(k)$ and let $\lambda \in k$. By definition of $\mathcal{P}_d(k)$, there exist coefficients $a_0,\dots,a_d \in k$ and $b_0,\dots,b_d \in k$ such that \begin{align*} p=\sum_{i=0}^{d} a_i x^i \end{align*} and \begin{align*} q=\sum_{i=0}^{d} b_i x^i. \end{align*} Addition in $k[x]$ is defined coefficientwise, so \begin{align*} p+q=\sum_{i=0}^{d} (a_i+b_i)x^i. \end{align*} Since $a_i+b_i \in k$ for every $i \in \{0,\dots,d\}$, we have $p+q \in \mathcal{P}_d(k)$. Scalar multiplication by elements of $k$ is also defined coefficientwise in $k[x]$, so \begin{align*} \lambda p=\sum_{i=0}^{d} (\lambda a_i)x^i. \end{align*} Since $\lambda a_i \in k$ for every $i \in \{0,\dots,d\}$, we have $\lambda p \in \mathcal{P}_d(k)$. The zero polynomial belongs to $\mathcal{P}_d(k)$ by taking all coefficients equal to $0_k$. Therefore $\mathcal{P}_d(k)$ is a $k$-subspace of the [vector space](/page/Vector%20Space) $k[x]$, hence is a vector space over $k$ under the inherited operations.[/step]

[guided]We want to prove that $\mathcal{P}_d(k)$ is a vector space without introducing any new operations. The operations are the usual polynomial addition and scalar multiplication from $k[x]$, so it is enough to verify that $\mathcal{P}_d(k)$ is closed under those operations and contains the zero polynomial. Let $p,q \in \mathcal{P}_d(k)$ and let $\lambda \in k$. The definition of $\mathcal{P}_d(k)$ means precisely that $p$ and $q$ can be written using only the monomials whose exponents are at most $d$. Thus there exist coefficients $a_0,\dots,a_d \in k$ and $b_0,\dots,b_d \in k$ such that \begin{align*} p=\sum_{i=0}^{d} a_i x^i \end{align*} and \begin{align*} q=\sum_{i=0}^{d} b_i x^i. \end{align*} Polynomial addition in $k[x]$ adds corresponding coefficients. Therefore \begin{align*} p+q=\sum_{i=0}^{d} (a_i+b_i)x^i. \end{align*} Because $k$ is a field, it is closed under addition, so each coefficient $a_i+b_i$ lies in $k$. Hence $p+q$ again has an expression using only $1,x,\dots,x^d$, and therefore $p+q \in \mathcal{P}_d(k)$. Similarly, scalar multiplication by $\lambda \in k$ multiplies each coefficient by $\lambda$: \begin{align*} \lambda p=\sum_{i=0}^{d} (\lambda a_i)x^i. \end{align*} Because $k$ is closed under multiplication, each $\lambda a_i$ lies in $k$, so $\lambda p \in \mathcal{P}_d(k)$. Finally, the zero polynomial is obtained by choosing every coefficient to be $0_k$: \begin{align*} 0=\sum_{i=0}^{d} 0_k x^i. \end{align*} Thus $\mathcal{P}_d(k)$ contains $0$, is closed under addition, and is closed under scalar multiplication. It follows that $\mathcal{P}_d(k)$ is a $k$-subspace of $k[x]$, and hence a vector space over $k$ with the inherited operations.[/guided]

[step:Prove that $1,x,\dots,x^d$ span $\mathcal{P}_d(k)$] Let $p \in \mathcal{P}_d(k)$. By definition, there exist $a_0,\dots,a_d \in k$ such that \begin{align*} p=\sum_{i=0}^{d} a_i x^i. \end{align*} Since $x^0=1$, this writes $p$ as a $k$-linear combination of the list $1,x,x^2,\dots,x^d$. Therefore $1,x,\dots,x^d$ span $\mathcal{P}_d(k)$ over $k$. [/step]

[step:Use uniqueness of coefficients to prove linear independence] Suppose $c_0,\dots,c_d \in k$ satisfy \begin{align*} \sum_{i=0}^{d} c_i x^i=0 \end{align*} in $k[x]$. By the defining uniqueness of coefficient representation in the polynomial ring $k[x]$, each coefficient of this polynomial is equal to the corresponding coefficient of the zero polynomial. Hence \begin{align*} c_i=0_k \end{align*} for every $i \in \{0,\dots,d\}$. Therefore the only $k$-linear relation among $1,x,\dots,x^d$ is the zero relation, so the list is linearly independent over $k$. [/step]

[step:Conclude the basis and dimension statement] The previous two steps show that the ordered list \begin{align*} 1,x,x^2,\dots,x^d \end{align*} spans $\mathcal{P}_d(k)$ and is linearly independent over $k$. Therefore it is a basis of $\mathcal{P}_d(k)$ over $k$. This basis contains exactly $d+1$ elements, namely one monomial $x^i$ for each integer $i \in \{0,\dots,d\}$. By the definition of dimension of a finite-dimensional vector space as the cardinality of a basis, we obtain \begin{align*} \dim_k \mathcal{P}_d(k)=d+1. \end{align*} This proves the theorem. [/step]