Alternating Fibonacci-like sequences are periodic? Define a Fibonacci-like sequence that depends on a
parameter $k \in \mathbb{N}$,
and alternates $\pm$:
\begin{eqnarray}
k=1 \;:\; f_1(n) &=& f_1(n-1)\\
k=2 \;:\; f_2(n) &=& f_2(n-1)-f_2(n-2)\\
k=3 \;:\; f_3(n) &=& f_3(n-1)-f_3(n-2)+f_3(n-3)\\
k=4 \;:\; f_4(n) &=& f_4(n-1)-f_4(n-2)+f_4(n-3)-f_4(n-4)\\
&\cdots&\\
k=k \;:\; f_k(n) &=& \Sigma_{i=1}^k (-1)^{i+1} f_k(n-i)
\end{eqnarray}
For any initial data specifying the values of
$f_k(n)$ for $n=0,1,2,\ldots,k{-}1$,
I claim the sequence becomes periodic, with period $(k+1)$ if $k$ is odd, and $2(k+1)$ if $k$ is even. 
For example, for $k=4$, and initial values
$$
\left(\; f_4(0),f_4(1),f_4(2),f_4(3) \;\right) = (1,2,3,4) \;,
$$
then $f_4(n)$, for $n=0,\ldots,20$ is:
$$
1, 2, 3, 4, 5, 2, -2, -3, -4, -5, -2, 2, 3, 4, 5, 2, -2, -3, -4, -5, -2 \;,
$$
with period $2(k+1)=10$.
For example, 
\begin{eqnarray}
f_4(5) &=& f_4(4)-f_4(3)+f_4(2)-f_4(1)\\
f_4(5) &=& 5-4+3-2\\
f_4(5) &=& 2 \;.
\end{eqnarray}
If instead we pin all initial values to $1$, so that
$$
\left( \; f_4(0),f_4(1),f_4(2),f_4(3) \; \right) = (1,1,1,1) \;,
$$
the resulting sequence is:
$$
1, 1, 1, 1, 1, 0, -1, -1, -1, -1, 0, 1, 1, 1, 1, 0, -1, -1, -1, -1, 0 \;,
$$
also period $10$.
My question:

Q. What is a proof of the claim that such alternating
  Fibonacci sequences $f_k(n)$ are periodic for any initial values?

I can prove that, e.g., $f_4(n)$ is periodic with period $10$, but only via induction for that specific $k{=}4$,
and initial conditions.
But if my claim is true, there should be a way to see that
all $f_k(n)$, regardless of initial values, are periodic with those odd/even periods $(k+1)$/$2(k+1)$.
 A: You can express $k=4$ for example using matrices
$$ \left( \begin{array}{c }
a_{n+1} \\
a_{n+2} \\
a_{n+3}  \\
a_{n+4} \end{array} \right) = \left( \begin{array}{cccc}
0 & 1 & 0 &0 \\
0 & 0 & 1 &0\\
0 & 0 & 0 & 1\\
-1& 1& -1& 1\end{array} \right) \left( \begin{array}{c }
a_n \\
a_{n+1} \\
a_{n+2}  \\
a_{n+3} \end{array} \right)$$
If you plug the matrix into Wolfram Alpha it says the eigenvalues $\lambda_1, \lambda_2,\lambda_3,\lambda_4$ are distinct and satisfy $\lambda_i^5 = \pm 1$. Prove it! Any sensible proof should work equally well for all even $k$.
Hence the matrix $A$ above has Jordan form $J$ and $A = S^{-1}J S$ for $$J = \text{diag}(\lambda_1, \lambda_2,\lambda_3,\lambda_4).$$
Then $$J^{10} = \text{diag}(\lambda_1^{10}, \lambda_2^{10},\lambda_3^{10},\lambda_4^{10}) = I$$ and so $$A^{10} = S^{-1}J^{10} S= S^{-1} S=I.$$ That means $$A^{10}(a_n,a_{n+1},a_{n+2},a_{n+3}) =  (a_n,a_{n+1},a_{n+2},a_{n+3}).$$ But by definition $$A^{10}(a_n,a_{n+1},a_{n+2},a_{n+3}) = (a_{n+10},a_{n+11},a_{n+12},a_{n+13}).$$ Thus the period is $10$.
For $k=5$ the eigenvalues are sixth roots of $6$. Prove it! Then do the same thing with $10$ replaced by $6$.
Bonus! One useful fact might be that if $\omega_0, \ldots, \omega_{k-1} $ are the $k$ distinct roots of unity then
$$\omega_0+ \omega_1 \ldots+ \omega_{k-1} =0.$$
To prove this recall $\omega_m = e^{m(2 \pi i/k)}$ so we can define
$$X = \omega_0+ \ldots+ \omega_{k-1}  = e^{0(2 \pi i/k)}+ e^{(2 \pi i/k)}+ \ldots+ e^{(k-1)(2 \pi i/k)}.$$
We claim $\omega_1 X = X$. Since $\omega_1 \ne 1$ this implies $X=0$. To prove it write
$$\omega_1 X = e^{(2 \pi i/k)}(e^{0(2 \pi i/k)}+ e^{(2 \pi i/k)}+ \ldots+ e^{(k-1)(2 \pi i/k)}) $$ $$ = e^{(2 \pi i/k)}+ e^{2(2 \pi i/k)}+ \ldots+ e^{(k-1)(2 \pi i/k)} + e^{k(2 \pi i/k)} $$ $$= \omega_1  +\omega_2 + \ldots + \omega_{k-1} + \omega_0 =X$$
