A number $x$ such that $x^n = x$ for every $n$ Is there proof fully algebraic that shows that $1$ is the only number that when put in the function $f(x) = x^n$, for $n = $ any number, the output is always $x$? 
This is pretty obvious, but you seem to prove it using the trivial knowledge . 
For example, $1^4 = 1, 1^{46373} = 1$, but $2^4 \neq 2$
 A: I'll provide an easy way:
We need the following to be true for all $n \in \mathbb N$
$$x^n=1$$ 
First let $x=1$, and we could check that $1^n=1$ for all $n \in \mathbb N$, so $x=1$ is one solution.
Do we have any other solutions? NO, to see this,
let $n=1$, the above equation needs to be true, and thus we get $x=1$.
So we are done.
EDIT
So the question now is changed to $x^n =x$, which makes things a bit different.
So now we want $$x^n=x$$
Still, easy to check $x=1,0$ are solutions for all $n > 0$, but notice that $0$ is not a solution when $n=0$, since $0^0$ is usually not defined. 
Then let $n=2$, $x^2=x$, thus we have to have $x = 1,0$
Thus the only solution is $1$.
EDIT
I'll take comment of @rachwieb, and try not to discuss $0^0$ here. But we let $x=0,1$ if we take all $n \in \mathbb N \setminus \{0\}$; and $x= 1$ if we take $n \in \mathbb Z$, because $0^{-1}$ is not defined.
A: Once we rule out $x=0$ (since we typically define $0^0=1$):
Assume $x^n = x$ for any $n$
then in particular $x^2 = x$
Divide both sides by $x$ (which is ok, since $x \not = 0$), and we get:
$x=1$
A: Be $x^n=x\quad \forall n\in \mathbb{N}$ then: 
Depending on how you define $0^0$ and if you include $0$ in $\mathbb{N}$ a solution might be $0$, as $0^n=0 \quad\forall n\in \mathbb{N}$
And if $x\neq0$, you can divide x from both sides of the equation $x=x^2$ and get that $x=1$ 
thus $0$ and $1$ are the only solutions.
A: Going off of Dr. Graubner's hint, we have:
$$f(x) - x = 0$$
$$x^n - x = 0$$
$$x(x^{n-1}-1) = 0$$.
Therefore, either $x$ or $x^{n-1}-1$ is equal to $0$.
$$x^{n-1}=1$$
$$(n-1)\ln(x) = 0$$
Now, either $(n-1)$ or $\ln(x)$ is $0$. Because $n$ is any number, we can't define it. Therefore:
$$\ln (x) = 0$$
$$x = 1$$
A: For positive integer $n$, write
$$f(x)=x^n=x^n-1+1=(1+x+\cdots+x^{n-1})(x-1)+1.$$
If $x > 1$, then $(1+x+\cdots +x^{n-1})(x-1) > n(x-1) $ so $f(x) > n(x+1)+1 > 1$.
If $0<x < 1$, then $(1+x+\cdots +x^{n-1})(x-1) < \frac{1}{1-x}(x-1) = -1$. Hence $f(x) < -1+1 = 0$.
A: I know this question was probably written without any thought of leaving $\mathbb R$ or $\mathbb C$, but I think it's a good opportunity to sharpen the question and show the big picture. The question apparently is, or could be:

How can I algebraically prove that if $x^n=x$ for all positive integers $n$, then $x=1$ or $x=0$?

(IMO the idea of discarding $0$ as a solution -- on the grounds of a particular chosen value of $0^0$ or otherwise -- is far less natural than just restricting the powers to be positive integers.)
There's actually no problem with immediately generalizing this to:

If $R$ is a ring with identity $1$ and no nonzero zero divisors, $x^2=x$ implies $x=1$ or $x=0$.

The proof has pretty much already been given above: if $x^2-x=0$, then $x(x-1)=0$, and by assumption either $x=0$ or $(x-1)=0$. This is just assuming the proposed problem for $n=2$, so of course requiring it for $n>2$ does no add anything new. (And assuming "$n$ is any number is a bit overkill too.)
But that is not quite the end of the story. You see, it was important here that the ring did not have nonzero zero divisors. In fact, there are very natural rings that don't have that property, and the proposed problem fails piteously. Namely:

There exists an infinite Boolean ring $R$, that is, a ring in which $x^2=x$ for all $x\in R$. In such a ring, $x^n=x$ for every element of the ring, for every positive integer $n$.

I chose that one to maximize the number of things being badly behaved, but of course it has nothing to do with cardinality. A Boolean ring with $4$ elements serves equally well to show that we could potentially have more than just $0$ and $1$ doing this.
