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...

We can develop the Linear Recurrence
$$
\begin{align}
a_n
&=\sum_{k=0}^n(-1)^k\binom{n-k}{k}2^{n-2k}\\
&=\sum_{k=0}^n(-1)^k\left[\binom{n-k-1}{k}+\binom{n-k-1}{k-1}\right]2^{n-2k}\\
&=2\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k}2^{n-2k-1}-\sum_{k=0}^{n-2}(-1)^k\binom{n-k-2}{k}2^{n-2k-2}\\[6pt]
&=2a_{n-1}-a_{n-2}
\end{align}
$$
which has the characteristic equation
$$
x^2-2x+1=0
$$
which has a double root at $x=1$. Thus, the solution has the form
$$
a_n=c_0\cdot1^n+c_1n\cdot1^n
$$
Since $a_0=1$ and $a_1=2$, we get
$$
a_n=n+1
$$
