Let $f: \Bbb{R} \to \Bbb{R}$ be the function defined by $f(t) = \det e^{tB}$. Suppose you already manage to establish that $f$ satisfies the ODE
\begin{align}
f'(t) &= \text{tr}(B)\cdot f(t)
\end{align}
Solving this is pretty easy; in fact this is the protoypical example of a separable ODE. The solution to this ODE is just
\begin{align}
f(t) &= f(0) \cdot e^{\text{tr}(B) \cdot t} \\
&= \det(I) \cdot e^{\text{tr}(tB)} \\
&= e^{\text{tr}(tB)}
\end{align}
And just to answer your other question, the solution to
\begin{align}
y'(t) = q(t)\cdot y(t)
\end{align}
subject to the initial conditions $y(t_0) = y_0$ is given by
\begin{align}
y(t) = y_0\exp\left( \int_{t_0}^t q(s) \, ds \right)
\end{align}
The "usual" way of justifying this is to divide both sides by $y(t)$ and then integrate:
\begin{align}
\dfrac{y'(t)}{y(t)} &= q(t) \\\\
\int \dfrac{y'(t)}{y(t)} \, dt &= \int q(t) \, dt \\
\ln \big( y(t) \big) &= \int q(t) \, dt
\end{align}
Now exponentiate both sides. I've purposely been sloppy with this method regarding constants of integration, because there is already a problem with this approach: it implicitly assumes $y(t)$ is no-where vanishing (because otherwise we cannot divide by $y(t)$). But hopefully you see where the solution "comes from". A more complete way which I like for solving this ODE is to just recognise that the formula for $y(t)$ I gave above is actually a solution, and then use a uniqueness argument to conclude there aren't anymore.
