The free associative algebra generated by one element/indeterminate is "the same thing as" the tensor algebra of a one-dimensional vector space, as far as I understand (1)(2)(3)(4)(5).
I'm not sure how much the choice of coefficients matters for the question; if it does matter, assume for the sake of simplicity that the coefficients are real numbers, $\mathbb{R}$.
I think this follows from the fact that the free semigroup generated by one element is commutative (and thus so is the free monoid generated by one element as well as the free group generated by one element). I.e., for the same reason the above facts are true, one can't really distinguish $X \otimes X \otimes X$ and $X^3$ in some sense (since there is only one indeterminate/unknown/variable, as long as we have associativity, expressions of the same length are indistinguishable, which implies commutativity).
I don't have a formal proof in mind and am not requesting a formal proof, since I am already fairly convinced that this is true, but would nevertheless appreciate a sanity check.
