Limit of this function as $x \rightarrow \infty$?

Question

I was interested in approximating $\log x$ without using the $\log$ function, specifically for values of $x \in [1, e]$, and I got this monster of a function:

$$\left( \frac{x-1}{e-1}\right)^{\frac{1}{ \left( \frac{1}{e-1} \right) \left( \frac{x-1}{e-1}\right)^{1/e} +1}}$$

It works as an approximation for $\log x$, very well I might add - the maximum error for $x \in [1, e]$ is ~$.00165...$. But this function is not equivalent to $\log x$, and that error $E \rightarrow \infty$ as $x \rightarrow \infty$, and this is because that function approaches a value. Or appears to, anyway. It could diverge to $-\infty$.

So what is the limit of this function? And how would I go about finding that? I would guess it approaches $1$, but I don't know how to prove it.

For readability's sake, I'll rewrite it like this:

$a = 1/e$

$b= 1/(e-1)$

$z= b(x-1)$ $$f(x) = z^{ \frac{1}{ b \ z^a +1}}$$