Prove If $A^{2014}$ is invertible, then $A$ is also invertible Use the associativity of matrix multiplication to prove that if $A^{2014}$ is invertible, then $A$ is also invertible.
Any help please?
 A: As help, consider $AB$ invertible, which means we can find $C$ with $(AB)C=I$ then can you use associativity to show $A$ is invertible? If $B=A$ what does this mean? Can you see how you could apply this to your problem by choosing $B$ appropriately? 
A: This is another way, even though Mark Bennet is more elementary and perfectly clear. 
I'm using the properties of the function "$det$" defined over the set of square matrix with value in the base field. 
By Cauchy - Binet formula $$det(AB)=det(A)det(B)$$ So in this problem, with an iterated reasoning, $$ det(A^{k}) = det(A)^{k} $$
Now, you need to prove that $det(A) \neq 0 \Leftrightarrow A$ is invertible. You can take a look here.
A: Let $B$ be the inverse of $A^{2014}$. This means
$$A^{2014}B = I$$
Based on the above equation, you want to find a $C$ such that
$$AC = I$$
So your task is to figure out what $C$ is from the first equation.
A: And yet another way of looking at this:
Saying that an element $a$ is invertible in a ring $R$ is equivalent to saying that multiplication by that element $R → R, x ↦ ax$ is invertible/bijective.
If $A^{2014}$ is invertible, then multiplying by $A^{2014}$ is bijective which is the $2014$-fold self-composition of multiplying by $A$. So multiplying by $A$ must be bijective as well, meaning $A$ is invertible.
A: Yet another way. $A$ is invertible iff $Ax=0$ implies $x=0$. Assume that $A^{2014}$ is invertible. Suppose that $Ax=0$. Then $A^{2014}x=A^{2013}Ax=0$, which implies $x=0$ because $A^{2014}$ is invertible. So $Ax=0$ implies $x=0$, which means $A$ is invertible.
