$f:\mathbb R \to \mathbb R, f(x)=e^x-x $. Show that for any $n\in \mathbb N, n \ge 2$ the equation $f(x) = n$ has an unique solution in $(0,\infty)$ $f:\mathbb R \to \mathbb R, f(x)=e^x-x$. Show that for any $n\in \mathbb N, n \ge 2$ the equation $f(x) = n$ has an unique solution in $(0,\infty)$.   
Before everything, I need to solve this using only 12th grade (calculus) knowledge, so keep that in mind please.  
I am trying to figure it out. Here is what I am thinking so far. My understanding of the question is that $f(x)$ must cross the axis $y=n$ just once. So, $f(x)$ must be strictly increasing over $(0,\infty)$ but I know this is not enough and I am not sure what the second condition is. I am thinking that if the value of $f(x)$ in any of the point $n$ is lower than $n$ it will never cross $y=n$ so how do I write the second condition? Is it $f(n) \ge n$? Is that all that I need to show? Because I can easily see both these conditions are true.
Don't be harsh please, this is a really different exercise than what I was used to and I am just trying to figure it out.
 A: $$f(x)=e^x-x \,\forall\,x\in R$$
function $f(x)$ is continuous and differentiable hence
$$f'(x)=e^x-1$$
This , $f(x)$ is increasing function for $x\ge 0$ and decreasing function for $x\le 0$.
$$. $$
Hence $f(x)$ will have minimum at $x=0$.
$$f(0)=1$$
This $f(x)\ge 1 \,\forall\,x\in R$
$$ $$
Range of $f(x)$ is $f(x)\in (\infty,1]$
$$. $$
Thus $f(x)=n$ has unique solution $x\gt 0$ also unique solution $x\lt 0$.
A: The derivative is $f'(x)=e^x-1>0$ for $x>0$ so $f$ is strictly increasing. Let $n\geq 2$ be given.
It's a good idea to review power series. The power series of $e^x$ is given by $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$. From this, it's easy to see that all terms are positive when $x>0$ so $e^x>\sum_{n=0}^2 \frac{x^n}{n!}=1+x+\frac{x^2}{2}$. Therefore, $e^x-x>1+\frac{x^2}{2}$.
We can then solve $1+\frac{x^2}{2}=n$ by $x=\sqrt{2(n-1)}$. In summary, when $x=\sqrt{2(n-1)}$, we have $e^x-x>1+\frac{x^2}{2}=n$.
Finally, by the intermediate value theorem, there exists $x\in(0,\sqrt{2(n-1)})$ such that $f(x)=n$ and since $f$ is strictly increasing, this solution is unique.
A: A bit of geometry.
Consider the intersection of $y_1=e^x$ with $y_2=x+n$, $x>0,$ $n\ge 2$.
$y_1=e^x = 1+x+x^2/2+...=$
$ y_2+(1-n+x^2/2+...)$.
1) For large $x$ the term in parentheses 
is $>0,$ i.e. $y_1$ is above $y_2$.
2) For very small $x >0$ the term in parentheses is negative, since $-1 \ge 1- n$ for $n\ge 2$ and the contribution from the $x$ terms is very small.
3) All functions are continuos: 
There is a zero in-between, i.e a point of intersection of $y_1$ and $y-2$.
4) Try to argue that there is only one zero.
A: To show the uniqueness say $a, b$ be two distinct roots with $a, b\in (0, \infty)$
So thee exists c with $a<c<b$ with $f'(c)=$
Here,  $f'(x)=e^x-1=0$ has only solution $x=0$ which is not an element of $(0, \infty)$. 
To prove,  existence of at least one root :
$f(0)=1-n<0$ and $f(n)>0$,  so now use intermediate value theorem. 
