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?

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.
