You can show that the amplitude has that form by thinking about the Feynman rules. I'll only discuss particles in the adjoint representation (this is suitable for supersymmetric theories), but this can be done more generally. Note that if you have particles transforming in the fundamental representation of $SU(N)$ then the amplitude does not have the form in your question.

Think about the three-gluon vertex. It contains a $f^{a b c}$ factor which, up to a constant, can be written as $\operatorname{tr}(T^a [T^b, T^c]) = \operatorname{tr}(T^a T^b T^c) - \operatorname{tr}(T^a T^c T^b)$. So this can be written as a combination of traces of products of Lie algebra generators. Now, think about joining together two such triple vertices. We have to compute quantities like $\operatorname{tr}(T^a T^b T^c) \operatorname{tr}(T^c T^d T^e)$, where the index $c$ is summed over.

Now, for $SU(N)$ we have that
$$
(T^a)_i^j (T^a)_k^l = \delta^j_k \delta^l_i - \frac 1 N \delta^i_j \delta^k_l,
$$ where we sum over the index $a$. For $U(N)$ the last term in the right hand side is absent. Using this, you'll see that the color factor for joining two triple vertices can also be written as a single trace. The same can done for the quartic vertex.

For a tree level scattering, your can do this recursively. You start with some triple vertex and you keep joining other vertices and by using the two identities above you can always rewrite the answer as a single trace.

At loop level, this doesn't work because you can get things like $\operatorname{tr}(T^a \cdots T^b T^c) \operatorname{tr}(T^c T^d \cdots T^b)$, with sum over $b$ and $c$. Using the $SU(N)$ identity you get a contribution $\operatorname{tr}(T^a \cdots T^b T^d \cdots T^b)$. Now using the identity again you get $\operatorname{tr}(T^a \cdots) \operatorname{tr}(T^d \cdots)$. In general, at $\ell$ loops you can have $\ell+1$ traces if you have a large enough number of particles.

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