Determinant of a sum of square matrices Let $$A=\begin{bmatrix}0&1&0&\cdots 0\\
0&0&1&\cdots 0\\
\vdots\\
0&0&0&\cdots 1 \\
1&1&1&\cdots1
 \end{bmatrix}_{n\times n}$$
i.e. it has ones above the main diagonal except for the last row and the last row has all ones.

I am trying to find $\det(A+A^2+\cdots+A^t)$ for $t\leq n $ $\tag{1}$

I have checked that for a few $n$, $\det(A)=\det(A^2)=\cdots=\pm 1$. But I am not sure how to prove that. Also determinant of sum of matrices is not equal to sum of determinant of those matrices, so I'm not sure how to find $(1)$? Any ideas?
**EDIT:**I realize that it is hard to find exact value of $(1)$ and so an upper bound for $(1)$ is also useful to me.

The characteristic polynomial of $A$ is $$\lambda^n-\lambda^{n-1}-\lambda^{n-2}-\ldots-1=0$$.  This has one root $\lambda_1\approx2$, and all the other roots are inside the unit circle.
$$det(I-A^t)=\prod (1-\lambda_i^t)\approx (\lambda_1^t)\lt2^t$$
