[proofplan] We construct the limit of an arbitrary finite diagram $D: J \to \mathcal C$ by encoding all cone equations as one equalizer. First take the product of the objects $D(j)$ over all objects $j \in \operatorname{Ob}(J)$, and the product of the objects $D(k)$ indexed by all morphisms $u: j \to k$ of $J$. Two maps from the first product to the second record, for each arrow $u: j \to k$, the two sides of the cone compatibility equation. The equalizer of these maps is then exactly the universal cone over $D$. [/proofplan] [step:Form the two finite products that encode vertices and arrows] Let $\operatorname{Ob}(J)$ denote the finite set of objects of $J$, and let $\operatorname{Mor}(J)$ denote the finite set of morphisms of $J$. For each morphism $u \in \operatorname{Mor}(J)$, write \begin{align*} u: j_u \to k_u \end{align*} for its domain object $j_u$ and codomain object $k_u$. Since $\mathcal C$ has finite products, choose a product object \begin{align*} P := \prod_{j \in \operatorname{Ob}(J)} D(j) \end{align*} with projection morphisms \begin{align*} p_j: P \to D(j) \end{align*} for each $j \in \operatorname{Ob}(J)$. Also choose a product object \begin{align*} Q := \prod_{u \in \operatorname{Mor}(J)} D(k_u) \end{align*} with projection morphisms \begin{align*} q_u: Q \to D(k_u) \end{align*} for each $u \in \operatorname{Mor}(J)$. [/step] [step:Define the parallel maps whose equalizer imposes cone compatibility] For each morphism $u: j_u \to k_u$ in $J$, the composite \begin{align*} D(u) \circ p_{j_u}: P \to D(k_u) \end{align*} is a morphism in $\mathcal C$, and $p_{k_u}: P \to D(k_u)$ is another morphism with the same codomain. By the universal property of the product $Q$, there exist unique morphisms \begin{align*} s: P \to Q, \qquad t: P \to Q \end{align*} such that, for every $u \in \operatorname{Mor}(J)$, \begin{align*} q_u \circ s &= D(u) \circ p_{j_u}, \\ q_u \circ t &= p_{k_u}. \end{align*} Let \begin{align*} e: L \to P \end{align*} be an equalizer of the parallel pair $s,t: P \to Q$. Thus \begin{align*} s \circ e = t \circ e, \end{align*} and for every object $X$ of $\mathcal C$ and every morphism $a: X \to P$ satisfying $s \circ a = t \circ a$, there exists a unique morphism $\bar a: X \to L$ such that \begin{align*} e \circ \bar a = a. \end{align*} [/step] [step:Extract a cone from the equalizer] For each object $j \in \operatorname{Ob}(J)$, define \begin{align*} \lambda_j := p_j \circ e: L \to D(j). \end{align*} We prove that the family $(\lambda_j)_{j \in \operatorname{Ob}(J)}$ is a cone over $D$. Let $u: j_u \to k_u$ be a morphism of $J$. Since $s \circ e = t \circ e$, composing both sides with the projection $q_u: Q \to D(k_u)$ gives \begin{align*} q_u \circ s \circ e &= q_u \circ t \circ e. \end{align*} Using the defining equations for $s$ and $t$, this becomes \begin{align*} D(u) \circ p_{j_u} \circ e &= p_{k_u} \circ e. \end{align*} By the definition of $\lambda_{j_u}$ and $\lambda_{k_u}$, this is exactly \begin{align*} D(u) \circ \lambda_{j_u} = \lambda_{k_u}. \end{align*} Thus $(L,(\lambda_j)_{j \in \operatorname{Ob}(J)})$ is a cone over $D$. [/step] [step:Construct the mediating morphism from any cone] Let $X$ be an object of $\mathcal C$, and let \begin{align*} \alpha_j: X \to D(j) \end{align*} for $j \in \operatorname{Ob}(J)$ be a cone over $D$. This means that for every morphism $u: j_u \to k_u$ in $J$, \begin{align*} D(u) \circ \alpha_{j_u} = \alpha_{k_u}. \end{align*} By the universal property of the product $P$, there exists a unique morphism \begin{align*} a: X \to P \end{align*} such that \begin{align*} p_j \circ a = \alpha_j \end{align*} for every $j \in \operatorname{Ob}(J)$. We show that $a$ equalizes $s$ and $t$. For every morphism $u: j_u \to k_u$ in $J$, \begin{align*} q_u \circ s \circ a &= D(u) \circ p_{j_u} \circ a \\ &= D(u) \circ \alpha_{j_u} \\ &= \alpha_{k_u} \\ &= p_{k_u} \circ a \\ &= q_u \circ t \circ a. \end{align*} Since the projections $(q_u)_{u \in \operatorname{Mor}(J)}$ jointly determine morphisms into the product $Q$, it follows that \begin{align*} s \circ a = t \circ a. \end{align*} Therefore, by the universal property of the equalizer $e: L \to P$, there exists a unique morphism \begin{align*} \bar a: X \to L \end{align*} such that \begin{align*} e \circ \bar a = a. \end{align*} For each $j \in \operatorname{Ob}(J)$, \begin{align*} \lambda_j \circ \bar a &= p_j \circ e \circ \bar a \\ &= p_j \circ a \\ &= \alpha_j. \end{align*} Thus $\bar a$ is a morphism from the cone $(X,(\alpha_j))$ to the cone $(L,(\lambda_j))$. [/step] [step:Prove uniqueness of the mediating morphism] Let $b: X \to L$ be any morphism satisfying \begin{align*} \lambda_j \circ b = \alpha_j \end{align*} for every $j \in \operatorname{Ob}(J)$. Then \begin{align*} p_j \circ e \circ b = \alpha_j \end{align*} for every $j \in \operatorname{Ob}(J)$. Since $a: X \to P$ is the unique morphism with $p_j \circ a = \alpha_j$ for every $j$, we get \begin{align*} e \circ b = a. \end{align*} The equalizer property of $e: L \to P$ gives uniqueness of the morphism $X \to L$ whose composite with $e$ is $a$. Since $\bar a$ also satisfies $e \circ \bar a = a$, we conclude that \begin{align*} b = \bar a. \end{align*} Therefore the cone $(L,(\lambda_j)_{j \in \operatorname{Ob}(J)})$ is terminal among cones over $D$, so it is a limit cone for $D$. Since $J$ and $D: J \to \mathcal C$ were arbitrary finite diagrams, $\mathcal C$ has all finite limits. [/step]