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 $2^4 \neq 2$

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.
