What is the best way to show that no positive powers of this matrix will be the identity matrix? 
Show that no positive power of the matrix $\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right)$ equals $I_2$.

I claim that given $A^{n}, a_{11} = 1$ and $a_{12} >0, \forall n \in \mathbb{N}$. This is the case for $n=1$ since $A^{1} = \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right)$ with $1=1$ and $1>0$.
Now assuming that $a_{11} = 1$ and $a_{12}>0$ for $A^{n}$ show that $A^{n+1} = A^{n}A = \left( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right)\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right) = \left( \begin{array}{cc} a_{11} & a_{11} + a_{12} \\ a_{21} & a_{21} + a_{22} \end{array} \right)$.
According to the assumption $a_{11} = 1$ and $a_{12}>0 \Rightarrow 1+a_{12} = a_{11}+a_{12}>0$. Taken together, this shows that $A^{n} \neq I_{2} \forall n\in \mathbb{N}$ since $a_{12}\neq0 = i_{12}$.
First of all, was my approach legitimate and done correctly? I suspect that I did not solve this problem as intended by the author (if at all!), could anyone explain the expected solution please? Thank you! 

Your solution seems OK to me. You can also find $A^n$ explicitly:
let $E=\left(\begin{array}{cc}0&1\\0&0\end{array}\right)$.
Then $A=I+E$ and $E^2=0$. So $(I+nE)(I+E)=I+(n+1)E$ and so, by induction, $A^n=I+nE\ne I$ for $n\ge1$.
