Free factor implies self-normalizing or trivial

Statement
Suppose $$G = H * K$$, i.e., $$G$$ is a free product of subgroups $$H$$ and $$K$$, so $$H$$ is a free factor of $$G$$. Suppose further that $$H$$ is a nontrivial group. Then, $$H$$ is a self-normalizing subgroup of $$G$$: $$N_G(H) = H$$.

Proof
Given: A free product $$G = H * K$$.

To prove: $$H$$ is self-normalizing in $$G$$.

Proof: Suppose $$H$$ is not self-normalizing in $$G$$. Pick any $$g \in N_G(H) \setminus H$$. Then we can write $$g$$ uniquely as an alternating product of elements from $$H$$ and $$K$$, with at least one letter from $$K$$. If the first letter for the unique expression for $$g$$ is $$h \in H$$, we can replace $$g$$ by $$h^{-1}g$$ to get a new element in $$N_G(H)$$ whose first letter is in $$K$$. Similarly, if the last letter is $$h' \in H$$, we can replace $$g$$ by $$gh'^{-1}$$ to get a new element in $$N_G(H)$$ whose last letter is in $$K$$. Thus, we can, without loss of generality, assume that:

$$g = k_1h_1 \dots k_{n-1}h_{n-1}k_n$$

where each $$h_i \in H, k_i \in K$$, and all elements are non-identity elements. Now, consider any non-identity element $$h \in H$$ (we can do this because $$H$$ is a nontrivial group). Clearly, the word $$ghg^{-1}$$ is also a reduced word, and since this word has length more than one, we see that $$ghg^{-1} \notin H$$. This contradicts the assumption that $$g \in N_G(H)$$, completing the proof.