<s>[INST] Difficulty Finding $A^k$ Let $A=
\begin{bmatrix}
1& -1 & 1\\ 
0 & 1 & 1 \\ 
0 & 0 &  1\\
\end{bmatrix}$. Compute $A^k$.
My attempt
I'm trying to compute $A^k$ using this approach as follows:
$$ 
A=I+N=
\begin{bmatrix}
1& 0 & 0\\ 
0 & 1 & 0 \\ 
0 & 0 &  1
\end{bmatrix}+
\begin{bmatrix}
0& -1 & 1\\ 
0 & 0 & 1 \\ 
0 & 0 &  0\\
\end{bmatrix}
$$
with
$$
N^2=
\begin{bmatrix}
0& 0 & -1\\ 
0 & 0 & 0 \\ 
0 & 0 &  0\\
\end{bmatrix}, \, \text{and} \, \,
N^3=
\begin{bmatrix}
0& 0 & 0\\ 
0 & 0 & 0 \\ 
0 & 0 &  0\\
\end{bmatrix}
$$
Then,
$$
A^2=(I+N)^2=I+2N+N^2, \\
A^3=(I+N)^3=I+3N+3N^2, \\
A^4=(I+N)^4=I+4N+6N^2, \\
A^5=(I+N)^5=I+5N+10N^2, \\
A^6=(I+N)^5=I+6N+15N^2,
$$
By induction, we can see $A^k=(I+N)^k=I+kN+f[k]N^2$. But, I couldn't figure out what $f[k]$ is. Any help?
 [/INST] How about this. Take the exponential function $e^{tA}$, where $t$ is some parameter
$$e^{tA}=\sum_{k=0}^\infty\frac{t^kA^k}{k!}=e^{t(I+N)}= e^{tI}e^{tN}=e^t\left[I+tN+\frac{(tN)^2}{2}\right]$$
where we used the matrix identity $e^{A+B}=e^Ae^B$ that is valid when 
matrices $A$ and $B$ commute.
Since the functions $t^k$ are linearly independent, we obtain
$$A^k=I+kN+\frac{k(k-1)}{2}N^2$$
</s>