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  "body": "\n## Proof\n\nWe triangularise $A$ over $\\C$ and compute both sides for a triangular matrix.\n\n1. If $\\mathbb{K} = \\R$, we view $A$ as a complex matrix. Its exponential is given by the same series, and its determinant and trace do not change. So we may assume that $A$ is a complex matrix.\n2. By [[triangularisation|0363JMQ42Y]] over $\\C$, there is an invertible complex matrix $P$ such that $U = P^{-1} A P$ is [[upper triangular|0MKABWDRNC]]. Write $U = (u_{ij})$.\n3. Let $S_N(X) = \\sum_{k=0}^{N} X^k / k!$. By induction on $k$, $U^k = P^{-1} A^k P$. So $S_N(U) = P^{-1} S_N(A) P$.\n4. Each entry of $P^{-1} S_N(A) P$ is a fixed linear combination of the entries of $S_N(A)$. By the [[definition of $e^A$|0K665H5V1S]] and the [[algebra of limits|07XBM3A9J5]], letting $N \\to \\infty$ in step 3 gives $e^U = P^{-1} e^A P$.\n5. Let $X, Y$ be upper triangular. For $i \u003e j$, each term $x_{il} y_{lj}$ of $(X Y)_{ij}$ needs $i \\le l \\le j$, which is impossible. So $X Y$ is upper triangular. The only nonzero term of $(X Y)_{ii}$ has $l = i$, so $(X Y)_{ii} = x_{ii} y_{ii}$.\n6. By step 5 and induction on $k$, $U^k$ is upper triangular with diagonal entries $u_{ii}^k$. This also holds for $k = 0$, with $U^0 = I_n$.\n7. So $S_N(U)$ is upper triangular, and its entry $(i, i)$ is $\\sum_{k=0}^{N} u_{ii}^k / k!$.\n8. Letting $N \\to \\infty$, $e^U$ is upper triangular. Its entry $(i, i)$ is the [[complex exponential|054VXHMCA3]] $e^{u_{ii}}$.\n9. By the [[determinant of a triangular matrix|0K1KKMFHYH]], $\\det e^U = e^{u_{11}} e^{u_{22}} \\cdots e^{u_{nn}}$.\n10. For complex numbers $z, w$, the $1 \\times 1$ matrices $(z)$ and $(w)$ commute, and their exponentials are $(e^z)$ and $(e^w)$. By the [[exponential of a sum of commuting matrices|04BD23PSK5]], $e^{z + w} = e^z e^w$.\n11. By step 10 and induction on $n$, $e^{u_{11}} \\cdots e^{u_{nn}} = e^{u_{11} + \\cdots + u_{nn}} = e^{\\tr U}$, where $\\tr$ is the [[trace|07PTPSZPME]].\n12. Since the [[trace of a product is symmetric|0Y4NF1DRKD]], $\\tr U = \\tr\\big(P^{-1} (A P)\\big) = \\tr\\big((A P) P^{-1}\\big) = \\tr A$.\n13. By step 4, $e^A = P e^U P^{-1}$. The [[determinant is multiplicative|04CPJ6AF00]], and complex numbers commute. So $\\det e^A = \\det e^U \\cdot \\det P \\det P^{-1} = \\det e^U \\det(P P^{-1})$.\n14. Since $P P^{-1} = I_n$ and [[$\\det I_n = 1$|0Z4BWARC38]], step 13 gives $\\det e^A = \\det e^U$.\n15. By steps 9, 11, 12 and 14, $\\det e^A = e^{\\tr U} = e^{\\tr A}$.\n\n## Idea\n\nBoth sides are unchanged when $A$ is replaced by a similar matrix. Over $\\C$, every matrix is similar to a triangular one. For a triangular matrix $U$, the exponential is triangular with diagonal entries $e^{u_{ii}}$, so its determinant is $\\prod_i e^{u_{ii}} = e^{\\tr U}$.\n",
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