Determinant of a matrix by induction

Base case. $1\times1$: trivial.

Inductive step. For an $n\times n$ upper triangular $T$ with diagonal $t_{11},\ldots,t_{nn}$, expand along the first column: only $t_{11}$ is nonzero (all entries below the diagonal are $0$). Cofactor is $(-1)^{1+1}=1$ times the $(n-1)\times(n-1)$ upper triangular submatrix with diagonal $t_{22},\ldots,t_{nn}$. By induction, this submatrix's determinant is $t_{22}\cdots t_{nn}$.

Conclusion. $\det T=t_{11}t_{22}\cdots t_{nn}$.