= Solution
A <simple Lie algebra> is a nonabelian <Lie algebra> whose only <Lie algebra ideals> are zero and the whole algebra. Assume $n\geq2$; $\mathfrak{sl}_1=0$ is not simple. We prove the assertion by the <matrix-unit extraction lemma for special linear ideals>, avoiding any classification theorem.
Let $I$ be a nonzero ideal in the <special linear Lie algebra> and $0\ne A\in I$. If an off-diagonal entry $A_{ji}$ is nonzero, two <commutators> with the <matrix unit> $E_{ij}$ give
$$
[E_{ij},[E_{ij},A]]=-2A_{ji}E_{ij}\in I,
$$
so $E_{ij}\in I$. If $A$ is diagonal and nonscalar, choose unequal diagonal entries; $[E_{ij},A]=(A_{jj}-A_{ii})E_{ij}$ again supplies a <matrix unit>. A nonzero scalar matrix cannot have zero <trace> over $\mathbb C$, so these cases cover every nonzero $A$.
Once $E_{ij}\in I$, its bracket with $E_{ji}$ gives $H=E_{ii}-E_{jj}$, and $[H,E_{ji}]=-2E_{ji}$ puts the reverse unit in $I$. Bracketing with the other off-diagonal <matrix units> generates every off-diagonal unit: for a third index $k$, $[E_{ij},E_{jk}]=E_{ik}$ and $[E_{ki},E_{ij}]=E_{kj}$; using the reverse unit fills the remaining row and column, then $[E_{ki},E_{i\ell}]=E_{k\ell}$ fills the others. Opposite pairs generate every trace-zero diagonal matrix. These vectors span $\mathfrak{sl}_n$, so
$$
\boxed{\mathfrak{sl}_n(\mathbb C)\text{ is simple for }n\geq2.}
$$
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