Product of binary matrices with binary eigenvalues Consider two binary matrices with obvious patterns:
$$
C=
\begin{bmatrix}
1 &0 &0 &0 &0 &0 &0\\
1 &0 &0 &0 &0 &0 &0\\
0 &1 &0 &0 &0 &0 &0\\
0 &1 &0 &0 &0 &0 &0\\
0 &0 &1 &0 &0 &0 &0\\
0 &0 &1 &0 &0 &0 &0\\
0 &0 &0 &1 &0 &0 &0
\end{bmatrix}
$$
and
$$
T=
\begin{bmatrix}
1 &1 &0 &0 &0 &0 &0\\
0 &1 &1 &0 &0 &0 &0\\
0 &0 &1 &1 &0 &0 &0\\
0 &0 &0 &1 &1 &0 &0\\
0 &0 &0 &0 &1 &1 &0\\
0 &0 &0 &0 &0 &1 &1\\
0 &0 &0 &0 &0 &0 &1
\end{bmatrix}
$$
The eigenvalues of the matrices $T^n C,n=0,1,2,3$ are zeros and consecutive powers of $2$ equal to $0,1,2,4$. I'd like to have a proof of the generalization of this fact for matrices of larger size with the same patterns.
Note, the entries of $T^n C$ in the left upper corner are zeros and binomial coefficients for the power $n+1$.
A motivation for this question is in 
Binary eigenvalues matrices and continued fractions

Some thoughts:
Note that $T = I + N$, where $I$ is the identity matrix and
$$
N = \pmatrix{0&1\\&0&1\\&&0&1\\&&&0&1\\&&&&0&1\\&&&&&0&1\\&&&&&&0}
$$
Notably, $N^7 = 0$. Because $NI = IN$, we can compute $T^n = (I + N)^n$ by binomial expansion.  That is, we have
$$
T^n = \binom n0 I + \binom n1 N + \cdots + \binom n6 N^6
$$
We can verify that $T^n$ is therefore the upper triangular Toeplitz matrix for which, in the notation of the linked wiki page, we have $a_{-k} = \binom nk$ whenever $0 \leq k \leq n$ and all other entries are $0$.
With that, we may compute 
$$
T^n C = \pmatrix{\binom n0 + \binom n1 & \binom n2 + \binom n3 & \binom n4 + \binom n5 & \binom n6 &0&0&0\\
\binom n0 & \binom n1 + \binom n2 & \binom n3 + \binom n4 & \binom n5 & 0&0&0\\
0 & \binom n0 + \binom n1 & \binom n2 + \binom n3 & \binom n4 & 0&0&0\\
0 & \binom n0 & \binom n1 + \binom n2 & \binom n3 & 0&0&0\\
0 & 0 & \binom n0 + \binom n1 & \binom n2 & 0&0&0\\
0 & 0 & \binom n0 & \binom n1 & 0&0&0\\
0 & 0 & 0 & \binom n0 & 0&0&0\\}
$$
