<s>[INST] Find $B$ if $B=A-{{1}\over{2}} A^2+{{1}\over{3}} A^3 -{{1}\over{4}} A^4+...$ 
Let $$
\ A=\begin{bmatrix} 0 & a & a^2 & a^3 \\ 0 & 0 & a & a^2 \\ 0 & 0 & 0 & a \\ 0 & 0 & 0 & 0 \end{bmatrix}
$$ and
  $B=A-{{1}\over{2}} A^2+{{1}\over{3}} A^3 -{{1}\over{4}} A^4+...$
$i)$ Find the matrix $B$
$ii)$ Prove that $A=B+ {{1}\over{2!}} B^2+ {{1}\over{3!}} B^3+...$

My attempt:
$i)$
I calculated $A^2$ by multiplying $A$ by itself, then foundnd $A^3$ by multiplying $A$ by $A^2$, ans so on.
Then I noted that $A^n=0$ for $n\geq 4$
$A^2= A.A=\begin{bmatrix} 0 & 0 & a^2 & 2a^3 \\ 0 & 0 & 0 & a^2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}
$
$A^3= A^2.A=\begin{bmatrix} 0 & 0 & 0 & a^3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}
$
$A^4=A^3.A =\begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}
$
$A^n=0$ for every $n\geq 4$, so:
$B= \begin{bmatrix} 0 & a & {{1}\over{2}} a^2 & {{1}\over {3}}a^3 \\ 0 & 0 & a & {{1}\over{2}} a^2 \\ 0 & 0 & 0 & a \\ 0 & 0 & 0 & 0 \end{bmatrix}
$
But is there any way easier than my way ?
And what about $(ii)$ ?
 [/INST] $B=\ln(I+A)$, and so $e^B=I+A\implies ii)$... 
You asked if there was a better way of doing it...  I believe the answer is to notice that you have the power series for $\ln$ and $\exp$ respectively;   and then use the fact (probably covered in your course) that those series can be applied to matrices.  See for instance . 
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