How do you show that: $1 + \frac{x}{1!} + \frac{x^2}{2!}+... = (1 + \frac{1}{1!} + \frac{1^2}{2!}+...)^x$ using elementary algebra? $e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!}+...$
$\rightarrow e = 1 + \frac{1}{1!} + \frac{1}{2!}+...$
$\rightarrow (e)^x = (1 + \frac{1}{1!} + \frac{1}{2!}+...)^x$
How do you show that:
$1 + \frac{x}{1!} + \frac{x^2}{2!}+... = (1 + \frac{1}{1!} + \frac{1}{2!}+...)^x$
using elementary algebra?
 A: The best we can do algebraically is to use “formal power series,” and then prove for integers $x.$
I’m going to gloss over a lot of steps here, this is just an outline.
We let $y$ be an indeterminate, and we define the ring of formal power series, written $\mathbb Q[[y]].$ (We could use $\mathbb R[[x]]$ instead, but $\mathbb Q[[x]]$ is more specific and, I feel, more algebraic.)
Then we prove that in this ring, if $$f(y)=\sum \frac {f_n}{n!}y^n,\\g(y)=\sum\frac{g_n}{n!}y^n$$ then $$f(y)g(y)=\sum \frac{h_n}{n!}y^n,$$
where $$h_n=\sum_{k=0}^n\binom nk f_kg_{n-k}\tag 1$$
Then we define $e(y)=\sum\frac1{n!}y^n,$ and prove by induction, using $(1),$ that, for any natural number $x,$
$$e(y)^x=\sum\frac{x^n}{n!}y^n=e(xy).$$
This proof is pretty much just the binomial theorem of $(x+1)^n.$
We can also show, using $(1),$ that $e(y)\cdot e(-y)=1.$ This follows from the binomial theorem on $(1+(-1))^n.$
We can use these to then show that, for any positive integer $q,$ $$e(\frac1qy)^q=e(y).$$ So $e(y/q)$ is a $q$th root of $e(y).$
But to move these results to facts about the real numbers, you need to make a statement about the meaning of convergence in $\mathbb R,$ and how these formal power series apply there.
This amounts to picking a sub-ring of $\mathbb Q[[y]]$ containing $e(y)$ and defining homomorphisms from that subring auto $\mathbb R.$
And even then, we’ve only shown it for $x$ rational. It is not even clear what $e^{\sqrt2}$ means, “algebraically.”

Side-note.
In $p$-adic numbers, which are, like the real numbers, an expansion of $\mathbb Q,$ $e(y)$ corresponds to a function which is defined for only a subset of values. In particular, $e(1)$ does not exist in $p$-adics, so, while $e(y)$ has properties of an exponential function where it is defined: $$e(y)^n=e(yn)\quad n\in\mathbb Z_p\\e(y_1)e(y_2)=e(y_1+y_2),$$ it has no $p$-adic number which is the equivalent of $e.$
So $e(y)$ is not intrinsically algebraically an exponent of some “number” $e.$ It turns out to be so in $\mathbb R,$ but we have to use information about convergence in $\mathbb R.$

The last is related to the fact that in $\mathbb Q[[y]],$ we have a general operation:
$$f\circ g$$ which is define only when $g$ is a multiple of $y.$ Which is to say, $g_0=0.$ This operation has the properties:
$$(f_1+ f_2)\circ g=f_1\circ g+f_2\circ g\\(f_1\cdot f_2)\circ g=(f_1\circ g)\cdot (f_2\circ g)\\(f\circ  g_1)\circ g_2= f\circ(g_1\circ g_2).$$
This lets us write things like $e(ay)$ when $a$ is rational.
Indeed, $\mathbb Q[[y]]$ has a very simple, if limited, form of convergence. A sequence $f_1,\dots,f_n,\dots$ converges to $0$ there if $f_i=y^{n_i}g_i$ for some sequence of power series $g_i$ and $n_i\to \infty.$
You get an addition property of $\circ:$
$$\lim_{n\to\infty} (f_n\circ g)=\left(\lim_{n\to\infty} f_n\right)\circ g$$ when the right side exists.
