Vtyjkyuk

Consider two binary matrices with obvious patterns:

C=
[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]

and

T=
[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]

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
$$
N = pmatrix0&1\&0&1\&&0&1\&&&0&1\&&&&0&1\&&&&&0&1\&&&&&&0
$$
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
$$
T^n C = pmatrixbinom n0 + binom n1 & binom n2 + binom n3 & binom n4 + binom n5 & binom n6 &0&0&0\
binom n0 & binom n1 + binom n2 & binom n3 + binom n4 & binom n5 & 0&0&0\
0 & binom n0 + binom n1 & binom n2 + binom n3 & binom n4 & 0&0&0\
0 & binom n0 & binom n1 + binom n2 & binom n3 & 0&0&0\
0 & 0 & binom n0 + binom n1 & binom n2 & 0&0&0\
0 & 0 & binom n0 & binom n1 & 0&0&0\
0 & 0 & 0 & binom n0 & 0&0&0\
$$

The left upper block of the matrix $T^n C$ can be conjugated to upper triangular one with the eigenvalues on diagonal by Pascal triangle matrix.

@Suvrit https://mathoverflow.net/questions/258284/is-the-matrix-left2m-choose-2j-i-right-i-j-12m-1-nonsingular