Proof of The Determinant of a Triangular Matrix is the Product of its Diagonal Entries

Let $S_{n}$ be the set of permutations of $[n]$, with the identity $\mathrm{id}$ of Permutations of an Initial Segment Form a Group under Composition, and let $S$ be the successor map of Natural Numbers.

Step 1: a permutation that never increases an index is the identity. Let $\sigma\in S_{n}$ satisfy $\sigma(i)\le i$ for every $i\in[n]$. We show by induction, using Principle of Induction for the Natural Numbers, the statement $P(m)$: if $m\le n$, then $\sigma(j)=j$ for every $j\in[m]$.

For $P(1)$: by claim 4 of Properties of the Order on the Natural Numbers we have $1\le\sigma(1)$, while $\sigma(1)\le1$ by hypothesis, so claim 2 of that lemma gives $\sigma(1)=1$; and $[1]=\{1\}$ by claim 2 of Basic Properties of Initial Segments of the Natural Numbers.

Assume $P(m)$ and let $S(m)\le n$. By claim 5 of Properties of the Order on the Natural Numbers we have $m<S(m)$, so $m\le n$ by claim 1 of that lemma, and the hypothesis $P(m)$ gives $\sigma(j)=j$ for every $j\in[m]$. Suppose $\sigma(S(m))\ne S(m)$. Since $\sigma(S(m))\le S(m)$, claim 5 of Properties of the Order on the Natural Numbers gives $\sigma(S(m))\le m$, so $j=\sigma(S(m))$ lies in $[m]$ and satisfies $\sigma(j)=j=\sigma(S(m))$. A permutation is injective by claim 1 of Injectivity, Composition, and Restriction of Bijections, so $j=S(m)$; but $j\le m$ and $m<S(m)$, which contradicts claims 1 and 2 of Properties of the Order on the Natural Numbers. Hence $\sigma(S(m))=S(m)$, and since $[S(m)]=[m]\cup\{S(m)\}$ by claim 3 of Basic Properties of Initial Segments of the Natural Numbers, this proves $P(S(m))$.

Applying $P(n)$ gives $\sigma(j)=j$ for every $j\in[n]$, that is, $\sigma=\mathrm{id}$.

Claim 1. Let $L$ be lower triangular and let $\sigma\in S_{n}$ with $\sigma\ne\mathrm{id}$. By Step 1 there is an $i\in[n]$ for which $\sigma(i)\le i$ fails, and then $i<\sigma(i)$ by claim 3 of Properties of the Order on the Natural Numbers. Hence $L_{i\,\sigma(i)}=0$, so claim 4 of Properties of Finite Products gives $\prod_{k=1}^{n}L_{k\,\sigma(k)}=0$ and the term of the determinant at $\sigma$ is $0$ by Zero Products and Elementary Identities in a Field.

Thus all terms outside the nonempty subset $\{\mathrm{id}\}$ of $S_{n}$ vanish, and claims 4 and 1 of Peeling, Splitting, and Interchange for Sums over a Finite Index Set, together with the value $\mathrm{sgn}(\mathrm{id})=1$ from claim 1 of The Sign of a Permutation is Multiplicative, give

$$\det L=\prod_{k=1}^{n}L_{k\,\mathrm{id}(k)}=\prod_{i=1}^{n}L_{ii}.$$

Claim 2. Let $L$ be upper triangular and let $L^{\top}$ be its transpose, so that $(L^{\top})_{il}=L_{li}$ for all $i,l\in[n]$. If $i<l$ then $L_{li}=0$, because $L$ is upper triangular, so $L^{\top}$ is lower triangular; and $(L^{\top})_{ii}=L_{ii}$ for every $i\in[n]$. Claim 1 applied to $L^{\top}$, together with claim 8 of Row Properties of the Determinant, gives

$$\det L=\det\bigl(L^{\top}\bigr)=\prod_{i=1}^{n}L_{ii}.$$