It seems quite intuitive when we say that some number $a$ is raised to a power $b$ where $a \in \mathbb{C} $ and $b \in \mathbb{Z}$ and can be expressed as $$a^b = a \times a \times a ... \text{($b$ times)}$$ Extending the argument such that $b \in \mathbb{R}$ then if $b$ is rational, it can be expressed in the form $\dfrac{p}{q}$ such that $ p,q \in \mathbb{Z}$ and $q \ne 0$ and $a^b$ is defined as $$a^{\frac{p}{q}} = \sqrt[q]{a^p}$$ If $b$ is irrational then $a^b$ is a transcendental number as stated by Gelfond- Schneider theorem ($a$ and $b$ are algebraic numbers). Agreed.
Now, here is the problem: What happens when $b$ is an imaginary number? What is an intuitive idea behind saying $i\theta$ times in the expression (I may be wrong in saying that) $$e^{i\theta} = e\times e\times e...\text{($i\theta$ times)} = \cos \theta + i\sin \theta$$ Yes, thats the Euler's theorem.
When the exponent is not a natural number (nonnegative integer), exponentiation is not repeated multiplication, any more than multiplying by $\lambda$ is repeated addition when $\lambda$ is not a natural number.
What we do in the case of both multiplication by $\lambda$ and raising to the $\lambda$ power is to give a new definition that agrees with the old definition when $\lambda\notin\Bbb N$, and that satisfies the same pleasant identities, namely $a^\lambda a^\mu=a^{\lambda+\mu}$, and $(ab)^\lambda=a^\lambda b^\lambda$, and $(a^\lambda)^\mu=a^{\lambda\mu}$.
Beyond all the above, there are special considerations when either the base is not a positive real or the exponent is not real. For instance, there’s no good consistent way of defining $(-1)^{1/4}$, though your taste may specify one value as more pleasing than some other one. All the worse if you tried to define $(-1)^\pi$. Any definition of these involves the log (natural) of $-1$, which has no special value that can be preferred over the others, except, again, by applying your taste.
The one good case is to define $e^z$ for $z$ being some complex number. There we have the wonderful formula already mentioned by @arctictern in his comment: $$ e^z=\sum_{k=0}^\infty\frac{z^k}{k!}\,. $$ Notice that this formula has nothing to do with multiplying $e$ to itself $z$ times. It’s just a formula that gives you $e\cdot e$ when $z=2$, and satisfies the other pleasant identities that I mentioned above.
And there you are.