One way to think of the complex plane $\mathbb{C}$ is to think of it as the real coordinate (or $xy$-) plane $\mathbb{R}^2$. In $\mathbb{R}^2$, we have two real-numbered axes (in particular, the $x$ and $y$ axes). An element of $\mathbb{R}^2$ is an ordered pair $(x, y)$, where $x$ and $y$ range over all the real numbers $\mathbb{R}$.
What I am about to say is not exactly rigorous, but I purposefully omit rigor in favor of your understanding.
Now, what happens if we decide to multiply $y$ by $i$, for all $y$ in $\mathbb{R}$? We obtain $i \mathbb{R}$, where each element is of the form of $i y$, where $i$ is of course defined as usual and $y$ is a real number. Consequently our vertical $y$-axis now becomes imaginary, and we obtain the complex plane $\mathbb{C}$. Note that just like the real plane $\mathbb{R}^2$, $\mathbb{C}$ is 2-dimensional. To recap, $\mathbb{C}$ has two axes, one real axis and one imaginary axis.
What does an element of $\mathbb{C}$ look like? An element $z$ of $\mathbb{C}$ has the form $a + bi$, where $a$ and $b$ are real numbers; we call $z$ a complex number. If we were to graph a complex number $z = a + bi$, we would graph it just like we would an ordered pair $(a, b)$ in $\mathbb{R}^2$. E.g., the number $z = 1 + i$ would correspond to the point $(1, 1)$; the number $z = 1$ would correspond to the point $(1, 0)$. This second example demonstrates that a real number is in fact a complex number as well.
I am not sure if you have seen linear algebra or not, but your statement in the second sentence of your first paragraph can be explained with such ideas. In the real plane $\mathbb{R}^2$, we have two directions; namely, the $x$ and $y$ directions. It turns out that we only need the number $x = 1$ to generate the entire $x$-axis. Similarly we may do the same with the $y$-axis. Together, we obtain the whole plane $\mathbb{R}^2$. $\mathbb{C}$ carries the same idea, except your vertical axis is generated by $i = 1\cdot i$ instead of just $1$.
To further address your question, recall $i$ is defined as $i = \sqrt{-1}$. Thus, $i^2 = -1$, $i^3 = -i$ and $i^4 = 1$. What have we accomplished here? We have obtained directions for the vertical and horizontal axes. From here we may generate all of $\mathbb{C}$.
I am not entirely sure how to answer your next few questions, but recall from above that $\mathbb{C}$ is 2-dimensional. Thus, $\mathbb{C}^2 = \mathbb{C} \times \mathbb{C}$ would be 4-dimensional. Generalizing this notion, $\mathbb{C}^n$ is $2n$-dimensional. Hence I do not think there is really an analogy to the $z$ axis if we are referring to the third axis in $\mathbb{R}^3$. The best analogy I can come up with are several-dimensional complex number systems such as $\mathbb{C}^2$.
We use the complex plane because it provides a very helpful visualization of complex numbers. The reason we can use it is because each complex number $z = a + bi$ is completely determined by the real numbers $a$ and $b$. In a way, thinking about a complex number as a point in $\mathbb{R}^2$ is just a different way of notating that number. There's no danger of confusion or ambiguity, because everyone can agree that a given point $(a,b)$ corresponds to the complex number $a + bi$, and vice versa. This is explained also in Atmos's answer.
I think it's worth mentioning that there is actually more than just this correspondence between the elements of each set, however.
We can also think about different operations that can be done with complex numbers and vectors in $\mathbb{R}^2$, and we can compare these. We know that complex numbers can be added together, as can vectors in $\mathbb{R}^2$. Complex numbers and vectors in $\mathbb{R}^2$ both add in the same way. Compare the following, for $a,b, c, d \in \mathbb{R}$
$$
(a + bi) + (c + di) = (a + c) + (b + d)i \\
(a,b) + (c,d) = (a + c, b + d)
$$
You can see that to add two complex numbers you can just think of each complex number as a point in $\mathbb{R}^2$ and add these points as you normally would. The result is the point in $\mathbb{R}^2$ corresponding to the sum of the complex numbers.
We can also scale complex numbers (multiply them by a real constant), and also scale vectors in $\mathbb{R}^2$. Again, this scaling behaves in the same way in $\mathbb{C}$ as it does in $\mathbb{R}^2$. Let $a,b,k \in \mathbb{R}$ and compare again
$$
k(a + bi) = ka + (kb)i \\
k(a,b) = (ka, kb)
$$
We see that to scale a complex number by a real constant, we can just think of the complex number as a point in $\mathbb{R}^2$, scale that point as we normally would, and the result is the point in $\mathbb{R}^2$ corresponding to the scaled complex number. If you'd like, the precise way of saying all of this is to say that $\mathbb{C}$ and $\mathbb{R}^2$ are isomorphic as $\mathbb{R}$-vector spaces. So you can be confident that the things you do with vectors (i.e. scalar multiplication, and vector addition) are the same in $\mathbb{C}$ and $\mathbb{R}^2$, so we are permitted to use this helpful visualization.
Edit: Here's an explanation of some extra behavior exhibited by $\mathbb{C}$ that is not compatible with the behavior of vectors in $\mathbb{R}^2$.
Above I did not explain the whole story of the multiplication operation. I explained how scaling by a real number works, but I did not mention what happens when you scale by an arbitrary complex number. Multiplication of complex numbers has a nice geometric interpretation when thinking of the elements of $\mathbb{C}$ as points on the plane. Each complex number $z = a + bi$ has associated to it a magnitude and an angle as measured counterclockwise from the positive $x$ axis. When multiplying complex numbers $z$ and $w$, the result is a new complex number with magnitude $|z|\cdot|w|$ and an angle that is the sum of the angles of $z$ and $w$. That is, multiplication of complex numbers corresponds to scaling and rotating on the complex plane. I explained a special case of this above when I talked about scaling by a real number. A real number has an angle of zero (it is on the $x$ axis), so multiplying by a real number only scales and does no rotation.
I mentioned that this operation is not compatible with a corresponding operation on $\mathbb{R}^2$. On one hand, this is the case because we are thinking of $\mathbb{R}^2$ as a vector space. And in a vector space, multiplication of two vectors need not be defined. So, this operation of multiplication is some sort of extra property possessed by the complex numbers that is not possessed by the vector space $\mathbb{R}^2$. This doesn't really change anything that I've already said, however. We can still think of complex numbers as points in $\mathbb{R}^2$. We just have to remember this extra rule for what happens to two points when we multiply them together.
And finally, some more detail if you still happen to be interested. We can define multiplication of elements in $\mathbb{R}^2$, and the natural way to do this is to say that for $a,b,c,d \in \mathbb{R}$
$$
(a,b) \cdot (c,d) = (ac, bc)
$$
This makes $\mathbb{R}^2$ into a ring. This way of multiplying elements, however, is not the same as the way we multiply elements in $\mathbb{C}$ because in $\mathbb{C}$ we have the relation that $i^2 = -1$, which changes things slightly. The precise way to state this difference is to say that $\mathbb{R}^2$ and $\mathbb{C}$ are not isomorphic as rings. Again though, this often doesn't matter to us, because in the situations where we want to think of $\mathbb{C}$ as $\mathbb{R}^2$ we are thinking of their compatibility as vector spaces, not as rings.
Best Answer
Euler's Formula is as follows;
$$e^{i\theta} = \cos\theta + i \sin \theta.$$
You can think of the number $1$ as $e^{\ln1}$ and hence of $1^i$ as $e^{i\ln1}$. But $\ln 1 = 0,$ so really you have $e^{0i} = e^0 = 1$, as I'm sure you're aware. Even if we look at this by feeding $0$ into Euler's Formula, we have
$$e^{0i} = \cos(0) + i\sin(0) = 1.$$