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...

Here is a generating function approach
$$
\begin{align}
\sum_{n=0}^\infty a_nx^n
&=\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k\binom{n-k}{k}2^{n-2k}x^n\tag{1}\\
&=\sum_{k=0}^\infty\sum_{n=k}^\infty(-1)^k\binom{n-k}{k}2^{n-2k}x^n\tag{2}\\
&=\sum_{k=0}^\infty\left(-\frac14\right)^k\sum_{n=k}^\infty\binom{n-k}{k}(2x)^n\tag{3}\\
&=\sum_{k=0}^\infty\left(-\frac14\right)^k\sum_{n=0}^\infty\binom{n}{k}(2x)^{n+k}\tag{4}\\
&=\sum_{k=0}^\infty\left(-\frac x2\right)^k\sum_{n=0}^\infty\binom{n}{k}(2x)^n\tag{5}\\
&=\sum_{k=0}^\infty\left(-\frac x2\right)^k\sum_{n=0}^\infty(-1)^{n-k}\binom{-k-1}{n-k}(2x)^n\tag{6}\\
&=\sum_{k=0}^\infty\left(-\frac x2\right)^k\sum_{n=0}^\infty(-1)^n\binom{-k-1}{n}(2x)^{n+k}\tag{7}\\
&=\sum_{k=0}^\infty\left(-x^2\right)^k\sum_{n=0}^\infty(-1)^n\binom{-k-1}{n}(2x)^n\tag{8}\\
&=\sum_{k=0}^\infty\left(-x^2\right)^k\frac1{(1-2x)^{k+1}}\tag{9}\\
&=\frac1{1-2x}\frac1{1+\frac{x^2}{1-2x}}\tag{10}\\
&=\frac1{(1-x)^2}\tag{11}\\
&=\sum_{k=0}^\infty(-1)^k\binom{-2}{k}x^k\tag{12}\\
&=\sum_{k=0}^\infty(k+1)x^k\tag{13}\\
\end{align}
$$
Explanation:
$\phantom{0}(2)$: change order of summation
$\phantom{0}(3)$: move $(-1)^k2^{-2k}=\left(-\frac14\right)^k$ out front
$\phantom{0}(4)$: substitute $n\mapsto n+k$
$\phantom{0}(5)$: move $(2x)^k$ out front
$\phantom{0}(6)$: $\binom{n}{k}=\binom{n}{n-k}=(-1)^{n-k}\binom{-k-1}{n-k}$ (see this answer)
$\phantom{0}(7)$: substitute $n\mapsto n+k$
$\phantom{0}(8)$: move $(2x)^k$ out front
$\phantom{0}(9)$: Binomial Theorem
$(10)$: sum of a geometric series
$(11)$: simplification
$(12)$: Binomial Theorem
$(13)$: $(-1)^k\binom{-2}{k}=\binom{k+1}{k}=\binom{k+1}{1}=k+1$
Equating the coefficients of $x^k$, we get $a_n=n+1$.
