Inverse of a triangular matrix Let $A$ be an $n \times n$ upper triangular matrix with integer entries and all diagonal entries are $1$

Prove / Disprove: $A^{-1}$ is not a power of $A$ unless $A$ is the identity matrix

I only know $A^{-1}$ exists and of course, it has all integer entries.
Any help?
 A: Hint: work out the $2 \times 2$ and $3 \times 3$ cases by hand (i.e., assume your matrix is 
$$
A = \begin{bmatrix}
1 & a \\ 0 & 1
\end{bmatrix}
$$
and its inverse is 
$$
A^{-1} = \begin{bmatrix}
1 & c \\ 0 & 1
\end{bmatrix}
$$
and see what you can conclude about $a$ and $c$. 
A: Denote $A=I+N$ where $N$ is some nilpotent non-zero upper triangular matrix and suppose $A^{-1}=(I+N)^k$.   
Then it should be $(I+N)^{k+1}=I$ and then with the use of of binomial formula $a_1N+a_2N^2+\dots+ a_{k+1}N^{k+1} =0 $    
$N(a_1I+a_2N +\dots+ a_{k+1}N^ k)  =0$     
but $B=a_1I+a_2N +\dots+ a_{k+1}N^ k$ is full rank matrix and $N$ in nonzero matrix so it is impossible to be $NB=0$.
So $N$ must be $0$.
A: Hint:  It is enough to prove that $A^m=I$ implies $A=I$.
Write $A=I+N$ with $N$ nilpotent. Then $A^m$ is easy to compute using the binomial theorem. Argue that $A^m=I$ implies $N=0$ by considering the diagonals of $N$. For instance, the secondary diagonal of $A^m$ is $m$ times the secondary diagonal of $N$.
