My definition of a profinite group is it's the inverse limit of a system of finite groups, $G=\varprojlim_{i\in I} G_i$.
But I can't see why $G=\varprojlim_{j\in J} G/N_j$, where $\{N_j\}_{j\in J}$ is the system of all open normal subgroups of $G$. I know each kernel of the projection $G\to G_i$ is an open normal subgroup of $G$, so $I$ is a subsystem of $J$. But I can't see why $I$ is cofinal in $J$. In order to show $I$ is cofinal in $J$, we need to show that for any $N_j\in \{N_j\}_{j\in J}$, there exists some $G_i$ with a homomorphism $G_i\to G/N_j$. But I have been stuck since then.
We need to show that every open normal subgroup of $G$ contains some $N_i := \ker(p_i : G \to G_i)$. In fact, every open neighborhood $1 \in U \subseteq G$ contains some $N_i$, so $(N_i)_{i \in I}$ is a basis of open neighborhoods of $1 \in G$.
Proof: Since $U$ is open and $1 \in U$, by the definition of the limit topology (=subspace topology of a product topology), there is a finite set $S \subseteq I$ of indices and open subsets $1 \in U_i \subseteq G_i$ for $i \in S$ with $\bigcap_{i \in S} p_i^{-1}(U_i) \subseteq U$. Then $\bigcap_{i \in S} N_i \subseteq U$. Let $k \in I$ be some index which is smaller* than all the indices in $S$. Then $N_k \subseteq N_i$ for all $i \in S$, hence $N_j \subseteq U$.
*This assumes that $i \leq j$ yields $G_i \to G_j$. When you have $G_j \to G_i$ instead (some authors do this), of course $k$ should be larger than all the indices in $S$.