power of a matrix+induction I am trying to solve the following problem:
Let $A$ the block form
$A= \begin{bmatrix}
B &C \\ 
 0&I 
\end{bmatrix}$
in which the blocks are $n \times n$. Prove that if $B-I$ is nonsingular, then, for $k \geq 1$,
$A^k = \begin{bmatrix}
B^k &(B^k-I)(B-I)^{-1}C \\ 
 0&I 
\end{bmatrix}$  (*)

I have tried:
Make a proof using induction
Case 1. $k=1$
$\begin{bmatrix}
B^1 &(B^1-I)(B-I)^{-1}C \\ 
 0&I 
\end{bmatrix}= \begin{bmatrix}
B &C \\ 
 0&I 
\end{bmatrix}=A = A^1$
This is because $(B-I)$ is nonsingular, this means that it has inverse, so $(B-I)(B-I)C = IC = C$
Case 2. $k > 1$
using the Caley-Hamilton theorem applied to $A$ I know that $C_A(A) = det\begin{bmatrix}
A-B &-C \\ 
 0&A-I 
\end{bmatrix}= A^2 - A(I+B)-IB = 0$
Then $A^2 = A(I+B)-IB$, so I can say that 
$A^k = A^{k-2}A^2 = A^{k-2}A(I+B)-IB$
the idea that i have is try to express (*) in this way but i don't know how
I appreciate any suggestion
Thank u
 A: Here's how I would complete the induction, assuming that (*), i.e ,
$A^k = \begin{bmatrix} B^k & (B^k - I)(B -I)^{-1}C \\ 0 & I \end{bmatrix}, \tag{1}$
holds for some integer $k \ge 1$.  First simply form $A^{k + 1} = AA^k$ with the matrices as given, obtaining
$A^{k + 1} = AA^k = \begin{bmatrix} B & C \\ 0 & I \end {bmatrix} \begin{bmatrix} B^k & (B^k - I)(B - I)^{-1}C \\ 0 & I \end {bmatrix}$
$= \begin{bmatrix} B^{k + 1} & B(B^k - I)(B - I)^{-1}C + C \\ 0 & I \end{bmatrix}; \tag{2}$
then observe that, for any matrix $B$, whether $B-I$ is invertible or not, we have
$(B - I) \sum_0^{k - 1} B^i = \sum_0^{k - 1} B^{i + 1} - \sum_0^{k - 1} B^i = B^k - I, \tag{3}$
for $k \ge 1$, whence if $(B - I)^{-1}$ exists we may write
$\sum_0^{k - 1} B^i = (B - I)^{-1}(B^k - I) = (B^k - I)(B - I)^{-1}. \tag{4}$
Using (4), we see that
$B(B^k - I)(B - I)^{-1}C + C = B(\sum_0^{k - 1} B^i)C + C = (\sum_0^{k - 1} B^{i + 1})C + C$
$= (\sum_0^k B^i)C = (B^{k + 1} - I)(B - I)^{-1}C, \tag{5}$
whence
$A^{k + 1} = \begin{bmatrix} B^{k + 1} & (B^{k + 1} - I)(B - I)^{-1}C \\ 0 & I \end{bmatrix}, \tag{6}$
which completes the induction and the requisite proof! QED.
Note Added in Edit; Friday 15 November 2013 9:57 PM PST:  Scrutiny of the preceding argument reveals that the case $B - I$ not invertible is swiftly addressed by minor variants the above assertions.  To wit, (4) indicates that we may write
$A^k = \begin{bmatrix} B^k & (\sum_0^{k - 1} B^i)C \\ 0 & I \end{bmatrix}, \tag{7}$
which in fact holds for any $B$, also seen by an even simpler inductive step:
$A^{k + 1} = AA^k = \begin{bmatrix} B & C \\ 0 & I \end {bmatrix} \begin{bmatrix} B^k & (\sum_0^{k - 1} B^i)C \\ 0 & I \end{bmatrix} = \begin{bmatrix} B^{k + 1} & (\sum_0^k B^i)C \\ 0 & I \end{bmatrix}; \tag{8}$
the intermediate algebraic operations are contained in equation (5) above.End:  Note Added in Edit.
Hope this helps!  Cheerio,
and as always,
Fiat Lux!!!
