Hint on proving $\sum_{k=0}^n {n \choose k} 2^k (-1)^k = (-1)^n$ I am trying to prove that this series : $$\sum_{k=0}^n {n \choose k}2^k (-1)^k = (-1)^n$$

However, i just honestly dont know where to begin. I tried expanding the LHS out and trying to use the fact that $n\choose k$ = $n \choose n-k$ however the $2^k$ in there is ruining things cancelling out nicely.

When i do a couple of examples on paper when n = 2 and n =3 , i seem to be getting things like: 

$n = 2: 1 + (4 - 4) = 1$

$n = 3: 1 + (6 - 12 + 8) = 1 - 2 = -1$

So i assume that $\sum_{k=1}^n {n \choose k} 2^k(-1)^k$ = 0 or -2 depending on even or odd respectively, Then once we add the first term of ${n \choose 0}2^0(-1)^0$ we'll get our 1 or -1

Im looking for a hint on how to approach this.

Thanks.
Hint: The binomial identity states
$$(a+b)^n=\sum_{k=0}^n\binom{n}{k}a^kb^{n-k}$$
For any real $a,b$ and nonnegative integer $n$. With $a=2$, $b=-1$, this almost becomes the sum you have, except that the exponent of $b$ is $n-k$ instead of $k$. But this is easily fixed, since the value of $(-1)^m$ only depends on the parity of $m$. Can you fill in the details?
Hint: do a binomial expansion on $(2-1)^n$