 Solution

Strictly, it is the infinitesimal generators, not every unitary matrix with unit determinant, that are traceless. For a continuous one-parameter family $U(ϵ)=\exp (iϵT)$, unitarity requires the generator $T$ to be Hermitian. The determinant condition givesDifferentiating at $ϵ=0$ yields $\mathrm{tr}T=0$. A Hermitian matrix has $n$ real diagonal parameters and $n(n-1)$ real off-diagonal parameters; one real trace constraint leaves $n^2-1$ independent generators. ThusThe identity matrix supplies a concrete counterexample to interpreting the claim as one about arbitrary group elements: $I_n\in SU(n)$ but $\mathrm{tr}{I}_n=n$. Also, $e^{i\mathrm{tr}T}=1$ at a single parameter value only fixes the trace modulo $2\pi$; the continuous neighbourhood is essential.