Show that $2^n-(n-1)2^{n-2}+\frac{(n-2)(n-3)}{2!}2^{n-4}-...=n+1$ 
If n is a positive integer I need to show that 
$2^n-(n-1)2^{n-2}+\frac{(n-2)(n-3)}{2!}2^{n-4}-...=n+1$

My guess: Somehow I need two equivalent binomial expression whose coefficients I need to compare.But which two binomial expressions? I know not!
P.S:Don't use Sterling Numbers or very high level maths...

I leave the judgement of uniqueness to the reader.
Let use consider the $r+1$th term of $$(2x^a-x^b)^{m-r}$$ which will be
$$\binom{m-r}r2^{m-2r}(-1)^rx^{am+r(b-2a)}$$
WLOG choose $b=2,a=1$
So, $\displaystyle\binom{m-r}r2^{m-2r}(-1)^r$ will be the coefficient of $x^m$ of  in $\displaystyle(2x-x^2)^{m-r}$
$$\implies\sum_{r=0}^{2r\le m}\binom{m-r}r2^{m-2r}(-1)^r$$ will be the coefficient of $x^m$ in the expansion of $$\sum_{r=0}^{2r\le m}(2x-x^2)^{m-r}$$
i.e., in the expansion of $$\sum_{r=0}^m(2x-x^2)^{m-r}=\sum_{u=0}^m(2x-x^2)^u=\dfrac{1-(2x-x^2)^{m+1}}{1-(2x-x^2)}=\{1-(2x-x^2)^{m+1}\}(1-x)^{-2}$$
i.e., in the expansion of $\displaystyle(1-x)^{-2}$
Now the coefficient in $x^n(n\ge0)$ in $\displaystyle(1-x)^{-2}$ (assuming the convergence) is
$$\dfrac{(-1)^n(-2)(-3)\cdots(-n)(-n-1)}{n!}=n+1$$
