How prove this function $f(x)=x!-x^n$ is injective Question:
For any positive integer $n$ such $n\neq 2^m-1, n\ge 2$, and function $f$ defined by 
$$f(x)=x!-x^n$$
show that :
$f:N\to Z$ is injective.
My idea: maybe for $x\neq y$ with $x,y\in N^{+}$, then this equation
$$x!-x^n=y!-y^n$$ has no solutions?
$$\Longleftrightarrow x!-y!=x^n-y^n$$
if $(x,y)=d$ then we let $x=dx_{1},y=dy_{1}$
then
$$(dx_{1})!-(dy_{1})!=d^n[(x_{1})!-(y_{1})!]$$
To solve this problem, maybe we can prove $f$ is monotonic? But I can't. It is said that this is from Mathematical olympiad problem.

 A: Note that the following proof does use the fact that $n\ne 2^m-1$ in the last line only.
The idea is to prove that when $x,y$ are big enough then the function $f(x)=x!-x^n$ is strictly increasing. The rest can be ruled out by considering some divisibility properties.
So let's suppose $x> y$ and $d_p(m)$ denotes the largest power of $p$ that divides $m.$  We have
$$x!-y!=x^n-y^n.$$
Take any prime divisor $p|y.$ Then, $p|\ LHS,$ so $p|x.$ Therefore,$$d_p(x!-y!)=d_p(y!)< \frac{y}{p}+\frac{y}{p^2}+...=\frac{y}{p-1}.$$
From the other hand $d_p(x^n-y^n)\ge n$ and we must have $n< \frac{y}{p-1}.$ In particular, $y>n.$
Suppose $p\ge 3,$ then $y>2n.$ This restriction is enough to establish monotonicity of $f.$ Indeed, it is enough to show that for $y>2n,$ $f(y+1)>f(y).$ In other words, we would like to show
$$(y+1)!-y!=y\cdot y!> (y+1)^n-y^n.$$
Using the inequality $y!\ge y^{\frac{y}{2}}$ that can be proved by induction or by grouping together $k$ and $y-k$ in the corresponding product, we can estimate
$$y\cdot y!\ge y\cdot y^n>y^{n+1}-y^n>(y+1)^n-y^n,$$
and the result follows.
So we are left to consider the case when  $p=2$ or $y=2^{\alpha}.$ Since $x$ is even, $d_2(x!-y!)=d_2(2^{\alpha}!)=2^{\alpha}-1.$ If $y=2^{\alpha}|x$ then $x\ge 2y> y+n$ and we can estimate
$$x!-y!=y!(x(x-1)...(y+1)-1)\ge (n+1)!(x(x-1)....(x-n)-1)\ge x^n$$ for $x\ge 2n,$
because $k(x-k)\ge x$ for $x\ge 2n$ and $1<k\le n.$
Hence, we have $d_2(x^n-y^n)=d_2(x^n)=\beta n$ and we mush have $\beta n=2^{\alpha}-1=y-1.$ 
Now we use the fact that $n\ne 2^m-1$ to conclude that $\beta\ne 1$ and thus $\beta\ge 3.$ This implies $y-1\ge 3n,$ $y>2n$ and this case was considered above.
A: for $x\neq y$ with $x,y\in N^{+}$, then this equation
$$x!-x^n=y!-y^n$$ has no solutions?
W.o.l.o.g. assume $ x < y $:
$$y^n-x^n=y!-x!$$
$$(y-x)(y^{n-1}+...+x^{n-1})=({y!}/{x!}-1)x!$$
Thus $ y^n $ has to be a multiple of $ x $...
