Theorem: Let $\mathbf{A}$ be category having all products and equalizers. Then $\mathbf{A}$ has all limits.

Proof: Let $D : \mathbf{I} \rightarrow \mathbf{A}$ be a diagram. We can define the following objects in $\mathbf{A}$:

$$O = \prod_{I \in \text{Ob}(\mathbf{I})} D(I) \qquad M = \prod_{u : J \rightarrow K \in \text{Mor}(\mathbf{I})} D(K)$$

Now we define maps $s, t : O \rightarrow M$.

$O$ is a product, and as such it comes with its projection maps $(\pi_I : O \rightarrow D(I))_{I \in \mathbf{I}}$. Likewise $M$ comes equipped with projections $\pi_u : M \rightarrow D(K)$ for each map $u : J \rightarrow K$. For any given $u : J \rightarrow K$, define a map $s_u: O \rightarrow D(K)$ by $s_u = Du \circ \pi_J$. Define $t_u : O \rightarrow D(K)$ by $t_u = \pi_K$. Finally, let $s$ be the unique morphism $O \rightarrow M$ such that $\pi_u \circ s = s_u$ for all morphisms $u$, and similarly for $t$.

Now, let $L \xrightarrow{p} O$ be the equalizer of $s$ and $t$. Write $p_I = \pi_I \circ p$. We claim that $\left( L \xrightarrow{p_I} D(I) \right)_{I \in \mathbf{I}}$ is precisely a limit cone over the whole diagram $D$.

First, let's verify that it is in fact a cone over $D$. Given $u : J \rightarrow K$ we require $D(u) \circ p_J = p_K$, or $D(u) \circ \pi_J \circ p = p_K$. Indeed, observe that $D(u) \circ \pi_J = s_u = \pi_u \circ s$, and then $\pi_u \circ s \circ p = \pi_u \circ t \circ p$ since $p$ is an equalizer. Finally, $\pi_u \circ t = t_u = \pi_K$, and $\pi_K \circ p = p_K$, giving the desired equality.

To show it unique, let $\left( A \xrightarrow{f_I} D(I) \right)_{I \in \mathbf{I}}$ be an arbitrary cone over $D$. There is a unique map $\varphi : A \rightarrow O$ such that $\pi_I \circ \varphi = f_I$. We claim that $s \circ \varphi = t \circ \varphi$. Indeed, consider that for some $u : J \rightarrow K$, we have $\pi_u \circ s \circ \varphi = s_u \circ \varphi = D(u) \circ \pi_J \circ \varphi = D(u) \circ f_J = f_K$. Meanwhile, $\pi_u \circ t \circ \varphi = t_u \circ \varphi = \pi_K \circ \varphi = f_K$. Since $\pi_u \circ (s \circ \varphi) = \pi_u \circ (t \circ \varphi)$ for all $u$, we have $s \circ \varphi = t \circ \varphi$. Then since $A$ equalizes $s$ and $t$, we have a unique map $\phi : A \rightarrow L$ with $p_I \circ \phi = f_I$. This is the desired unique map, making $L$ the limit of the diagram $D$.