A category with finite products is monoidal

Let $X\otimes Y:=X\times Y$ and let the unit object be $1$. On morphisms, define $(f,g)\mapsto f\times g$ by the universal property of the binary product. Because product pairings are unique, identities and compositions are preserved componentwise, so $\times :\mathcal{C}\times \mathcal{C}\to \mathcal{C}$ is a bifunctor.

Let ${\lambda }_X:(1\times X)\to X$ and ${\rho }_X:(X\times 1)\to X$ be the canonical product projections. If ${!}_X:X\to 1$ is the unique map from [L2], then the inverse of ${\lambda }_X$ is the pairing $⟨{!}_X,1_X⟩:X\to 1\times X$, and the inverse of ${\rho }_X$ is $⟨1_X,{!}_X⟩:X\to X\times 1$. Hence $\lambda$ and $\rho$ are natural isomorphisms.

For objects $X,Y,Z$, let ${\alpha }_{X,Y,Z}:((X\times Y)\times Z)\to X\times (Y\times Z)$ be the unique arrow whose three composite projections are the obvious first, second, and third projections. Define ${\alpha }_{X,Y,Z}^{-1}$ similarly. By the same uniqueness argument, these arrows are inverse and natural.

Both sides of the pentagon are arrows from $(((W\times X)\times Y)\times Z)$ to $W\times (X\times (Y\times Z))$. They have the same four composites with the terminally iterated product projections, so [L1] makes them equal. The same argument on the two projections from $(1\times (X\times Y))$ to $X$ and $Y$ shows the triangle equation.

Therefore $(\mathcal{C},\times ,1,\alpha ,\lambda ,\rho )$ is a monoidal category. The coproduct statement is proved by the same construction with pairings replaced by copairings and terminal replaced by initial, using the dual clauses already present in Products and coproducts as limits and colimits of discrete diagrams, including their existence-and-uniqueness equations and Initial object, terminal object, and zero object.