Can you put a direct sum in the exponent to represent the multiplicity of irreducible representations? It is well known that one can decompose tensor product spaces into direct sums and that sometimes the components of the direct sums are not unique. Taking Wikipedia's example, the addition of three spin-$1/2$ particles yields a single spin-$3/2$ particle and two spin-$1/2$ particles:
$$\mathbf{2}\otimes\mathbf{2}\otimes\mathbf{2}=\mathbf{4}\oplus\mathbf{2}\oplus\mathbf{2}.$$ Can this be written as $$\mathbf{2}\otimes\mathbf{2}\otimes\mathbf{2}=\mathbf{4}\oplus\mathbf{2}^{\oplus2}?$$ And, in general, can we write direct sums of the form
$$\bigoplus_k \mathbf{k}^{\oplus d_k},$$ where $d_k$ is the multiplicity associated with dimension $k$? I have seen this written as $$\bigoplus_k d_k\mathbf{k},$$ for example on the same Wikipedia page and in the references cited there, but that seems more confusing, especially if we try to decompose spin operators as $$\mathbf{J}=\bigoplus_k d_k \mathbf{J}_k \quad \mathrm{versus}\quad \mathbf{J}=\bigoplus_k \mathbf{J}_k^{\oplus d_k}.$$
 A: You evidently missed my point about trivial arithmetic checks on the Kronecker product reduction, in the convention where representations are labelled by their dimensionality, used here for SU(2) (in contrast to the atomic physics convention) and flavor SU(3) and GUTs in particle physics.
In this convention,
$$\mathbf{2}\otimes\mathbf{2}\otimes\mathbf{2}\otimes\mathbf{2}=\mathbf{5}\oplus\mathbf{3}\oplus\mathbf{3}\oplus\mathbf{3}\oplus\mathbf{1}\oplus\mathbf{1}.$$
Converting irreps to numbers (their dimensionality!), and direct products into products and direct sums to sums, you get a numerical grade-school arithmetic check; in this case, you get 16=16.
That is,
16×16 matrices reduce to a 5×5 matrix block, three 3×3 blocks, and two 1×1 singlet entries/blocks, upon a Clebsch orthogonal transformation.  The dimensionality of the reducible representation is then borne out by inspection!
As you see in your WP reference, the established condensed combinatoric convention for the above is, instead,
$$
\mathbf{5}\oplus 3\times\mathbf{3} \oplus 2\times\mathbf{1}, 
$$
comporting with this arithmetic check, albeit clumsily; it is mostly useful to the Catalan triangle algorithm.
In your stillborn scheme,  by contrast,  $ \mathbf{5}\oplus\mathbf{3}^{\oplus 3} \oplus\mathbf{1}^{\oplus 2}$, you'd get 6+27=33, instead, ensuring confusion for your hapless reader.
You are free to devise any conventions you like, but caveat lector.
NB Unfriendliness  to the reader is routine in this field, so, instead of using the above SU(2) expression, in atomic physics they use irrep labels j for d=2j+1 above,
$$\mathbf{1/2}\otimes\mathbf{1/2}\otimes\mathbf{1/2}\otimes\mathbf{1/2}=\mathbf{2}\oplus\mathbf{1}\oplus\mathbf{1}\oplus\mathbf{1}\oplus\mathbf{0}\oplus\mathbf{0},$$
which forfeits the trivial arithmetic check, straightforward to perform in your head, however.
I have the distinct impression you have not fully mastered the irrep coproduct (spin-composition in QM) which dictates this notation and makes the expressions so simple and natural,
$$
\Delta(J_z)= J_z\otimes \mathbb{1}_2\otimes \mathbb{1}_2\otimes \mathbb{1}_2 +\mathbb{1}_2\otimes J_z\otimes  \mathbb{1}_2\otimes \mathbb{1}_2 \\  + \mathbb{1}_2\otimes \mathbb{1}_2\otimes J_z\otimes \mathbb{1}_2+   
  \mathbb{1}_2\otimes \mathbb{1}_2\otimes \mathbb{1}_2 \otimes J_z.$$ It acts on each subspace simultaneously ("synchronized swimming") in the suitable irrep. The structure is additive, not multiplicative, so multiplexing is a multiplication, not an exponentiation.
