= Solution
For a <Lie group> $G$ with identity $e$, its <Lie algebra> as a set is the <tangent space> $T_eG$:
$$
\mathfrak g=\{\gamma'(0):\gamma:(-\varepsilon,\varepsilon)\to G\text{ smooth},\ \gamma(0)=e\}.
$$
The ambient embedding identifies these tangent vectors with vectors in $\mathbb R^N$. For a matrix <Lie group>, the identity is $I$ and the tangent vectors are matrices.
If $\gamma(t)\in\mathrm{SL}_n$ with $\gamma(0)=I$, put $A=\gamma'(0)$. The <determinant> expansion $\det(I+tA+o(t))=1+t\operatorname{tr}A+o(t)$ follows directly from its permutation formula: to first order only the diagonal entries contribute. Since $\det\gamma(t)=1$, differentiation gives $\operatorname{tr}A=0$. Thus
$$
\boxed{\operatorname{Lie}(\mathrm{SL}_n)\subseteq\mathfrak{sl}_n.}
$$
In fact equality holds: if $\operatorname{tr}A=0$, then $\exp(tA)$ has <determinant> $\exp(t\operatorname{tr}A)=1$ and derivative $A$ at zero.
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