Prove the function $f(x)=1-x-(1-\frac{a}{n}x)^n$ has a real solution between $(0,1]$ Consider the function $f(x)=1-x-(1-\frac{a}{n}x)^n$ where $a$ and $n$ are both constants such that $\frac{a}{n}\in[0,1]$ ( it represents some probability indeed). One obvious solution is $x=0$. I want to prove that such a function has a real solution between $(0,1]$ and this solution has the form that dependening on $a$ and independent of $n$. The only naive technique that came to mind is to use Rolle's theorem. However, it seems that it is not a reasonable approach.
============================================================
I want to add some clarifications to my question. $n\ge 2$ is a positive integer. As a very special case, for $n=2$, we have $f(x)= 1-x-(1-\frac{a}{2}x)^2$. When $\frac{a}{2}>\frac{1}{2}$, this function has a solution $x=\frac{4(a-1)}{a^2}$. Since I set $n=2$, I don't know how to interpret the solution is independent of $n$.
This question is one part of my homework, I believe I have my $f(x)$ correct. Thanks for your time and sorry for any inconvenience.
 A: The solution cannot be independent of $n$, even if $n$ is a positive integer. To see this, you can assume to the contrary and note that you get the same answer as when you take the limit of solutions as $n \to \infty$. But when $a$ is rational, the limit as $n \to \infty$ is the solution of $1 - x - e^{-ax} = 0$. Any non-zero solution to this is not algebraic, even though the solution is algebraic for any finite $n$.
A: This is false. Set $a=1$ and $n=2$ which gives $\frac{a}{n}=\frac{1}{2}\in[0,1]$ and you'll have:
$$f(x) = 1-x - (1- \frac{1}{2}x)^2$$
But simple algebraic manipulation shows that 
$$f(x) = 1-x - (1- \frac{1}{2}x)^2=-\frac{x^2}{4}$$
which has only one root at $x=0$ with multiplicity $2$.
A: (As made in a comment to the question $a = 0,n$ leads to no real solutions in $(0,1)$, and will thus be ignored.)
The general answer is no, if $a$ is small enough. We can show that there exist values of $a$ close to 0 in which for $ x \in (0, 1]$,
$$ 1-x < (1 - \frac{a}{n}x ) ^n.$$
The reason is that when $x=0$, we have equality. By differentiation, we just need to get a value of $a$ where
$$-1 < (1 - \frac{a}{n} x)^{n-1} (-a)$$
for all $x \in (0,1)$. We are already given that $n \geq 2 > 1$, so observe that for $a <1 $ works since
$$ 1 > (1 - \frac{a}{n} x )^{n-1} > (1 - \frac{a}{n} x )^{n-1} a.$$

This same argument shows that a root exists when
$$\lim_{x \rightarrow 0} (1 - \frac{a}{n} x)^{n-1} (-a) < -1$$

some1.new's answer is a special case of this.
Robjohn gives a great answer without using calculus which shows $a<1$ would never work for $n \geq 1$.
A: This cannot be true. Bernoulli's Inequality says that $\left(1-\frac anx\right)^n\ge1-ax$ when $\frac anx\le1$ and $n\ge1$.
Therefore, if $a\lt1$ and $0\lt x\le1$
$$
1-x-\left(1-\frac anx\right)^n\le(a-1)x\lt0
$$
If $a=1$, Beroulli's Inequality also says that for $0\lt x\le1$ and $n\ge2$, $\left(1-\frac1nx\right)^n\gt1-x$.
Therefore,
$$
1-x-\left(1-\frac1nx\right)^n\lt0
$$
The only way with $a\le1$ that the equation can be true is if $x=0$ or $a=1$ and $n=1$.
