Prove that $\exp : \mathbb{C} \to \mathbb{C}$ is periodic using as little geometry as possible. The proof attempt below is long, please consider it to be an attachment of sorts. The main content of the question is the introduction paragraph, and the proof attempt is my attempt to answer my question myself.
I know it's possible to define $\sin$ and $\cos$ (with either real or complex arguments) using $\exp$ and it's possible to define $\pi$ as the period of $\exp$ multiplied by $\frac{1}{2i}$ .
However, I'm curious how you would convince a skeptic that $\exp$ is periodic to begin with without appealing to the things that you are trying to build on top of it.
In particular, I don't want to use the identity $\exp(it) = \cos(t) + i \sin(t) $ and knowledge that $\sin$ and $\cos$ are both periodic.
I think I found a way to do it using the fact that $t \mapsto (-1) + \exp(it)$ has at least one positive root. I'm also using the fact that $\exp$ is multiplicative without proving it, largely out of space concerns.
I got somewhat stuck attempting to prove that $c$ (defined below) does in fact send at least one positive real to $1$ ... and ended up appealing to some notion of angle indirectly expressed in terms of the inner product.
I'm curious what the right way is to prove that $\exp$ is periodic, if you're planning to make that a foundational result that other things are built on top of.
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Proof Attempt
First, a word on notation.
$$ \exp{x} \stackrel{\text{def}}{=} \sum_{k = 0}^{\infty} \frac{x^k}{k!} $$
$$ c(z) \stackrel{\text{def}}{=} \exp(iz) = \cos(z) + i\sin(z) $$
Let $\langle \cdot, \cdot \rangle$ denote the inner product of two complex numbers as vectors in $\mathbb{R}^2$ .
$$ \langle x, y \rangle \stackrel{\text{def}}{=} \begin{bmatrix} \Re(x) \\ \Im(x) \end{bmatrix} \cdot \begin{bmatrix} \Re(y) \\ \Im(y) \end{bmatrix} = \Re(x\overline{y}) $$
$$ \theta(x, y) \stackrel{\text{def}}{=} \frac{\langle x, y \rangle}{|x||y|} $$
Lemma #1 $c$ sends reals to the unit circle.
$$ |c(t)| = 1 \; \forall t \in \mathbb{R}$$
The negation leads to a contradiction.
$$ |c(t)| \ne 1 \tag{NG} $$
$$ |c(t)|^2 \ne 1^2 $$
$$ c(t)\overline{c(t)} \ne 1 $$
$$ \exp(it)\overline{\exp(it)} \ne 1 $$
$$ \exp(it + \overline{it}) \ne 1 $$
$$ \exp(it - it) \ne 1 $$
$$ \exp(0) \ne 1 $$
$$ 1 \ne 1 $$
$$ \bot $$
Lemma #2 the angle between $c(a)$ and $c(1+a)$ is constant for real $a$
$$ \theta(c(0), c(1)) = \theta(c(a), c(1+a)) \;\forall a \in \mathbb{R} $$
The negation is a contradiction.
$$ \theta(c(0), c(1)) \ne \theta(c(a), c(1+a)) \tag{NG} $$
$$ \frac{\langle c(0), c(1)\rangle}{|c(0)||c(1)|} \ne \frac{\langle c(a), c(1+a)\rangle}{|c(a)||c(1+a)|} $$
$$ \langle c(0), c(1)\rangle \ne \langle c(a), c(1+a)\rangle $$
$$ \langle 1, c(1)\rangle \ne \langle c(a), c(1+a)\rangle $$
$$ \Re\left(\overline{c(1)}\right) \ne \Re\left(c(a)\overline{c(1+a)}\right) $$
$$ 2 \times \Re\left(\overline{c(1)}\right) \ne 2 \times \Re\left(c(a)\overline{c(1+a)}\right) $$
$$ \overline{c(1)} + \overline{\overline{c(1)}} \ne c(a)\overline{c(1+a)} + \overline{c(a)\overline{c(1+a)}} $$
$$ \overline{c(1)} + c(1) \ne c(a)\overline{c(1+a)} + \overline{c(a)}c(1+a) $$
$$ c(1) + c(-1) \ne c(a)c(-1-a) + c(-a)c(1+a) $$
$$ c(1) + c(-1) \ne c(-1) + c(1) $$
$$ c(1) + c(-1) \ne c(1) + c(-1) $$
$$ \bot $$
Lemma #3 The angle between $c(0)$ and $c(1)$ is non-zero.
$$ \theta(c(0), c(1)) \ne 0 $$
The negation leads to a contradiction.
$$ \theta(c(0), c(1)) = 0 \tag{NG} $$
$$ \frac{\langle c(0), c(1) \rangle}{|c(0)||c(1)|} = 0$$
$$ \langle c(0), c(1) \rangle = 0 $$
$$ 2 \times \langle c(0), c(1) \rangle = 0 $$
$$ 2 \times \Re\left( c(0)\overline{c(1)} \right) = 0 $$
$$ c(0)\overline{c(1)} + \overline{c(0)\overline{c(1)}} = 0 $$
$$ c(0)\overline{c(1)} + \overline{c(0)}c(1) = 0 $$
$$ c(0)c(-1) + c(0)c(1) = 0 $$
$$ c(-1) + c(1) = 0 $$
$$ \left(\sum_{k=0}^{\infty} \frac{(-i)^k}{k!}\right) +\sum_{k=0}^{\infty} \frac{(i)^k}{k!} = 0$$
$$ \sum_{k=0}^{\infty} \frac{(-i)^k + (i)^k}{k!} = 0 $$
$$ \sum_{k=0}^{\infty} \frac{(i^{-1})^k + (i)^k}{k!} = 0 $$
$$ \sum_{k=0}^{\infty} \frac{(i)^{(-k)} + (i)^k}{k!} = 0 $$
$$ \sum_{k=0}^{\infty} \frac{\overline{(i)^{k}} + (i)^k}{k!} = 0 $$
$$ \sum_{k=0}^{\infty} \frac{2\times \Re\left((i)^{k}\right)}{k!} = 0 $$
$$ 2 \times \sum_{k=0}^{\infty} [k \equiv_2 0] \frac{(-1)^{\frac{k}{2}}}{k!} = 0 $$
$$ 2 \times \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} = 0 $$
$$ \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} = 0 $$
But the LHS is positive because it's a monotonically decreasing alternating series whose first term is positive.
$$ \bot $$
Lemma #4 There is at least one positive real $b$ such that $c(b) = c(0) = 1 $
Okay, this step I'm a little stuck on. Intuitively, my argument is that $c$ has the intermediate value property and increasing $c$'s argument by $1$ moves you the same angular distance along the unit circle. Therefore, if you keep going in the positive direction you'll eventually pass $c(0)$ (which is $1$) again.
Lemma #5 Let $b$ be a positive number such that $c(b) = 1$, $\exp(bi) = 1$
$$ \exp(b) = 1 $$
The negation leads to a contradiction.
$$ \exp(bi) \ne 1 \tag{NG} $$
$$ c(i^{(-1)}bi) \ne 1 $$
$$ c(b) \ne 1 $$
$$ \bot $$
Lemma #6 $\exp$ is periodic with period $bi$ .
Note that there is no guarantee that $b$ is the smallest positive value for which this holds.
$$ \exp(z+bi) = \exp(z) \;\forall z \in \mathbb{C} $$
The negation leads to a contradiction.
$$ \exp(z+bi) \ne \exp(z) \;\exists z \in \mathbb{C} $$
$$ \exp(z)\exp(bi) \ne \exp(z) \exists z \in \mathbb{C} $$
$$ \exp(z)\times 1 \ne \exp(z) \exists z \in \mathbb{C} $$
$$ \exp(z) \ne \exp(z) \exists z \in \mathbb{C} $$
$$ \bot $$
 A: One can use non elementary tools (that do not rely on $\pi$ or geometry !) to prove this as well, in the following way. 
You already checked that $|c(x)| =1$ for all $x\in\mathbb R$. Now let $S=\{z\in S^1\mid z\in c(\mathbb R)\}$. 
The point will be to prove that it is open and closed in $S^1$, and to prove that $S^1$ is connected. 
First to show that it is open : this is where I would use differential geometry, although there are probably more elementary solutions. 
Indeed one can check that $S^1$ is a $1$-dimensional submanifold of $\mathbb R^2$ from its defining equation, and one can check that $c$ has a nonzero (therefore invertible, because we're in dimension $1$) differential by just computing it and knowing that $\exp$ doesn't vanish, so that $c$ is a local diffeomorphism. It follows that it has an open image. 
Now to show that the image is closed : this is where I would use differential calculus, although again there are probably more elementary solutions. 
Indeed $c$ satisfies the following differential equation $(E)\,\, y' =iy$ from $\mathbb R\to \mathbb C$. Moreover, for any $z\in S^1$, by taking $zc$ we see that $(E)$ has a solution with $z$ in its image.
Let $U:= zc(\mathbb R)$ which is an open neighbourhood of $z$. If $z\in \overline{S}$ then there is $y\in \mathbb R$ with $c(y) \in U$, say $c(y) = zc(x)$ for some $x\in \mathbb R$. By uniqueness of solutions to degree $1$ differential equations given an initial condition, it follows that $t\mapsto zc(t+x-y)$ and $c$ agree on a neighbourhood of $y$ and so we can glue them to get a solution $d$ to the equation which agrees with $c$ on $0$ and has $z$ in its image (this is very badly written, hopefully you'll understand what I mean - if you don't I can come back and make this bit more precise)
But that solution must be $c$ ! Therefore $z$ is in the image of $c$. So $S$ is closed. 
Therefore $S$ is clopen in $S^1$.
Finally, to show that $S^1$ is connected, note that the usual retraction from algebraic topology $\mathbb R^2\setminus \{0\}\to S^1$, $x\mapsto \frac{x}{||x||}$, and the proof of its continuity don't rely in any way on $\pi$. It's quite easy to show that $\mathbb R^2\setminus\{0\}$ is path-connected, therefore so is $S^1$. 
Since $0\in S$, it follows that $S=S^1$ so that $c:\mathbb R\to S^1$ is surjective. Let $x$ be such that $c(x) = -1$ (in complex notation). Then $c(2x) = 1$ and $2x\neq 0$, so $c$ is $2x$-periodic; now by standard arguments we find a smallest period $T$ which we call $2\pi$. And the rest, as they say, is history. 
This solution has the disadvantage of using heavier artillery (differential calculus, differential geometry), but the end result is that it is more conceptual, and somewhat cleaner than the alternative computational solution offered by LutzL (which, to be honest, I didn't read in detail) - although their solution is also instructive, in that it is quite elementary and readable with less background. 
It is very likely that the "local diffeomorphism" bit of the argument can be made more elementary, so the "$S$ is open" bit should be doable elementarily. I'm less certain about the "$S$ is closed" bit. As to connectedness of $S^1$, I mentioned algebraic topology, but in fact of course the argument is completely elementary. 
