Previous two posts:
Update: The first link only verifies continuity on $\mathbb R$, and so continuity cannot be used for complex heights. The limit of ${}^{z+n}a$ as $n\to\infty$ converges to the same value for all $z\in\mathbb C$ since $[\ln({}^\infty a)]^{z+n}\to0$.
Let $D_a$ be the points where the below definition converges.
I believe I have managed to prove that for $a\in(1,e^{1/e})$ and $z\in D_a$,
$${}^za=\lim_{n\to\infty}\log_a^{\circ n}({}^\infty a-({}^\infty a-{}^na)[\ln({}^\infty a)]^z)$$
is the unique tetration under the conditions that
${}^0a=1$
$\displaystyle{}^{z+1}a=a^{({}^za)}$ for all $z\in D_a$
$\displaystyle\lim_{n\to\infty}\frac{{}^\infty a-{}^{n+z}a}{{}^\infty a-{}^na}=[\ln({}^\infty a)]^z$ as a limit over $n\in\mathbb N$ for all $z\in D_a$.
Here is my attempted proof:
Let $b\pm\epsilon$ refer to a value within $\epsilon$ of $b$ for simplicity.
From $(3)$ we know that for all $\epsilon>0$, there exists $N$ s.t. for all $n>N$,
$$\frac{{}^\infty a-{}^{n+z}a}{{}^\infty a-{}^na}=[\ln({}^\infty a)]^z\pm\epsilon$$
By continuity of exponentiation, this can be rewritten in terms of another $\bar\epsilon>0$:
$$\frac{{}^\infty a-{}^{n+z}a}{{}^\infty a-{}^na}=[\ln({}^\infty a)]^{z\pm\bar\epsilon}$$
Solving for ${}^{n+z}a$ gives
$${}^{n+z}a={}^\infty a-({}^\infty a-{}^na)([\ln({}^\infty a)]^{z\pm\bar\epsilon})$$
Logging $n$ times and applying $(2)$ gives
$${}^za=\log_a^{\circ n}({}^\infty a-({}^\infty a-{}^na)([\ln({}^\infty a)]^{z\pm\bar\epsilon}))$$
Taking the limit as $n\to\infty$ gives us
$${}^za=\lim_{n\to\infty}\log_a^{\circ n}({}^\infty a-({}^\infty a-{}^na)([\ln({}^\infty a)]^{z\pm\bar\epsilon}))$$
However, from the first link, we know that this limit converges to a continuous function on $D_a$. So for this to be true for all $\bar\epsilon>0$, we must have
$${}^za=\lim_{n\to\infty}\log_a^{\circ n}({}^\infty a-({}^\infty a-{}^na)([\ln({}^\infty a)]^z))$$
Note also that the definitions of ${}^\infty a$ and ${}^na$ for natural $n$ in the above are defined by $(1)$ and $(2)$.
