[proofplan]
We first prove the identity for upper triangular matrices by tracking what happens to diagonal entries under powers and then under the matrix exponential series. In that case, $\exp T$ remains upper triangular and its diagonal entries are the scalar exponentials $\exp(T_{ii})$, so its determinant is the product of those entries. For a general complex matrix, Schur triangularisation writes it as a similarity transform of an upper triangular matrix. Similarity invariance of trace, determinant, and the matrix exponential then reduces the general case to the triangular case.
[/proofplan]
[step:Compute the exponential of an upper triangular matrix on the diagonal]
Let $T\in M(n,\mathbb C)$ be upper triangular, meaning that $T_{ij}=0$ whenever $i>j$. Let $I_n$ denote the $n\times n$ identity matrix. We claim that for every integer $m\ge 0$, the matrix $T^m$ is upper triangular and
\begin{align*}
(T^m)_{ii}=T_{ii}^m
\end{align*}
for every $i\in\{1,\dots,n\}$, where $T^0=I_n$.
For $m=0$, this holds because $T^0=I_n$ is upper triangular and $(I_n)_{ii}=1=T_{ii}^0$. Suppose the claim holds for some $m\ge 0$. If $i>j$, then
\begin{align*}
(T^{m+1})_{ij}=\sum_{k=1}^{n}(T^m)_{ik}T_{kj}.
\end{align*}
For each summand, either $k<i$, in which case $(T^m)_{ik}=0$ because $T^m$ is upper triangular, or $k\ge i>j$, in which case $T_{kj}=0$ because $T$ is upper triangular. Hence $(T^{m+1})_{ij}=0$, so $T^{m+1}$ is upper triangular. For the diagonal entry,
\begin{align*}
(T^{m+1})_{ii}=\sum_{k=1}^{n}(T^m)_{ik}T_{ki}.
\end{align*}
If $k<i$, then $(T^m)_{ik}=0$, and if $k>i$, then $T_{ki}=0$. Thus only the term $k=i$ remains, giving
\begin{align*}
(T^{m+1})_{ii}=(T^m)_{ii}T_{ii}=T_{ii}^{m+1}.
\end{align*}
The claim follows by induction.
Define the matrix exponential by the absolutely convergent series
\begin{align*}
\exp T=\sum_{m=0}^{\infty}\frac{T^m}{m!}.
\end{align*}
For $i>j$, every partial sum has $(i,j)$-entry equal to $0$, so $(\exp T)_{ij}=0$. Thus $\exp T$ is upper triangular. For each $i\in\{1,\dots,n\}$, entrywise convergence of the matrix exponential series gives
\begin{align*}
(\exp T)_{ii}=\sum_{m=0}^{\infty}\frac{(T^m)_{ii}}{m!}=\sum_{m=0}^{\infty}\frac{T_{ii}^m}{m!}=\exp(T_{ii}).
\end{align*}
[guided]
Let $T\in M(n,\mathbb C)$ be upper triangular, so $T_{ij}=0$ whenever $i>j$. The point of this step is to prove that the exponential series cannot create entries below the diagonal, and that the diagonal behaves exactly as in the scalar exponential.
First prove the corresponding statement for powers. We claim that for every integer $m\ge 0$, the matrix $T^m$ is upper triangular and
\begin{align*}
(T^m)_{ii}=T_{ii}^m
\end{align*}
for every $i\in\{1,\dots,n\}$. For $m=0$, the identity $T^0=I_n$ gives an upper triangular matrix, and its diagonal entries satisfy $(I_n)_{ii}=1=T_{ii}^0$.
Assume the claim holds for a fixed $m\ge 0$. We compute entries of $T^{m+1}=T^mT$. If $i>j$, then
\begin{align*}
(T^{m+1})_{ij}=\sum_{k=1}^{n}(T^m)_{ik}T_{kj}.
\end{align*}
We inspect each summand. If $k<i$, then $(T^m)_{ik}=0$ because the inductive hypothesis says $T^m$ is upper triangular. If $k\ge i$, then $k>j$, since $i>j$, and therefore $T_{kj}=0$ because $T$ is upper triangular. Thus every summand is zero, so $(T^{m+1})_{ij}=0$. This proves that $T^{m+1}$ is upper triangular.
Now take $i=j$ in the same product formula:
\begin{align*}
(T^{m+1})_{ii}=\sum_{k=1}^{n}(T^m)_{ik}T_{ki}.
\end{align*}
If $k<i$, then $(T^m)_{ik}=0$. If $k>i$, then $T_{ki}=0$. The only possible nonzero summand is $k=i$, and hence
\begin{align*}
(T^{m+1})_{ii}=(T^m)_{ii}T_{ii}.
\end{align*}
Using the inductive hypothesis on the diagonal entry gives
\begin{align*}
(T^{m+1})_{ii}=T_{ii}^mT_{ii}=T_{ii}^{m+1}.
\end{align*}
This completes the induction.
Now use the matrix exponential series
\begin{align*}
\exp T=\sum_{m=0}^{\infty}\frac{T^m}{m!}.
\end{align*}
Because every $T^m$ is upper triangular, every partial sum of this series is upper triangular. Entrywise convergence then implies that the limit $\exp T$ is also upper triangular. For a diagonal entry, the previous power computation gives
\begin{align*}
(\exp T)_{ii}=\sum_{m=0}^{\infty}\frac{(T^m)_{ii}}{m!}.
\end{align*}
Substituting $(T^m)_{ii}=T_{ii}^m$ yields
\begin{align*}
(\exp T)_{ii}=\sum_{m=0}^{\infty}\frac{T_{ii}^m}{m!}=\exp(T_{ii}).
\end{align*}
Thus the matrix exponential of an upper triangular matrix is upper triangular, and its diagonal entries are the scalar exponentials of the diagonal entries of the original matrix.
[/guided]
[/step]
[step:Evaluate the determinant in the upper triangular case]
Since $\exp T$ is upper triangular, its determinant is the product of its diagonal entries. Using the diagonal computation from the previous step,
\begin{align*}
\det(\exp T)=\prod_{i=1}^{n}(\exp T)_{ii}=\prod_{i=1}^{n}\exp(T_{ii}).
\end{align*}
The scalar exponential satisfies the finite product law, so
\begin{align*}
\prod_{i=1}^{n}\exp(T_{ii})=\exp\left(\sum_{i=1}^{n}T_{ii}\right).
\end{align*}
By definition of trace,
\begin{align*}
\sum_{i=1}^{n}T_{ii}=\operatorname{tr}T.
\end{align*}
Therefore
\begin{align*}
\det(\exp T)=\exp(\operatorname{tr}T)
\end{align*}
for every upper triangular $T\in M(n,\mathbb C)$.
[/step]
[step:Triangularise the arbitrary complex matrix]
Let $X\in M(n,\mathbb C)$. By Schur triangularisation over $\mathbb C$, there exist a unitary matrix $Q$, meaning a complex matrix with $Q^*Q=I_n$, and an upper triangular matrix $T\in M(n,\mathbb C)$ such that
\begin{align*}
X=QTQ^{-1}.
\end{align*}
The hypotheses of Schur triangularisation are satisfied because $X$ is a complex square matrix.
By the conjugation invariance of the exponential applied with $A=Q$,
\begin{align*}
\exp X=\exp(QTQ^{-1})=Q(\exp T)Q^{-1}.
\end{align*}
Taking determinants and using multiplicativity of the determinant together with $\det(Q^{-1})=\det(Q)^{-1}$ gives
\begin{align*}
\det(\exp X)=\det(Q)\det(\exp T)\det(Q^{-1})=\det(\exp T).
\end{align*}
[Trace is invariant under similarity](/theorems/7811): using the cyclic identity $\operatorname{tr}(AB)=\operatorname{tr}(BA)$ for square matrices of the same size,
\begin{align*}
\operatorname{tr}X=\operatorname{tr}(QTQ^{-1})=\operatorname{tr}(TQ^{-1}Q)=\operatorname{tr}T.
\end{align*}
[/step]
[step:Transfer the triangular identity back to the original matrix]
The triangular case applied to the upper triangular matrix $T$ gives
\begin{align*}
\det(\exp T)=\exp(\operatorname{tr}T).
\end{align*}
Using the determinant and trace identities obtained from the Schur triangularisation step,
\begin{align*}
\det(\exp X)=\det(\exp T)=\exp(\operatorname{tr}T)=\exp(\operatorname{tr}X).
\end{align*}
This is the desired formula for the original matrix $X\in M(n,\mathbb C)$.
[/step]
