On Scalar products — What is wrong with this argument?

Question

Suppose $|k\rangle =\exp(k \hat O^\dagger)|0\rangle$ where $c_k\in \mathbb C$ and $|0\rangle$ is normalized.

I wish to evaluate $\langle a|b\rangle$.

Here is what I think, but the result is not right, could someone please tell me what is wrong with it?

It is not hard to show that $\hat O|k\rangle=k|k\rangle$

So $\langle a|b\rangle= \langle 0|(1+a^*\hat O+{1\over 2}(a^*)^2(\hat O)^2+...)|b\rangle$

$=\exp(a^*b)\langle 0|b\rangle$

$=\exp(a^*b)\langle 0| \exp(b \hat O^\dagger)|0\rangle$

$=\exp(a^*b)\langle \exp(b^* \hat O)(0)|0\rangle$

because I think $\langle \exp(b^* \hat O)(0)|=\exp(b \hat O^\dagger)|0\rangle^\dagger$ ---$(*)$

$=\exp(a^*b)\langle 0|0\rangle$

$=\exp(a^*b)$

This cannot be right. Since for one thing, it is strange that $\langle 0|b\rangle=\langle 0|0\rangle$.

Also, I expected that the "vectors" should be orthonormal!

Thank you!