Perhaps it makes more sense if you rewrite the inequality and state the proof in steps:
Premise: $\text{det}A>1.$
Suppose, for the sake of contradiction, that there is NOT an eigenvalue $\lambda_i > 1$. In other words, assume that for all $\lambda_i,\;1\leq i \leq n,\;\lambda_i \leq 1$.
Then if all $|\lambda_j | \leq 1$, their product: $|\lambda_1 ...\lambda_n| = |\text{det}A| \leq 1$.
This is a contradiction to the premise: $\text{det} A > 1$.
Hence, the supposition is false, and we conclude that if it's the case that $\text{det} A >1$, then it must be the case that at least one eigenvalue must be greater than 1.
EDIT:
It is a contradiction because if we take as true that $\text{det} A > 1,$ but at the same time suppose that all eigenvalues $|\lambda_i| \leq 1$, then their product must be less than or equal to one, but their product is equal to $\text{det} (A)$, so $\text{det} (A)$ would then be $\leq 1$ This contradicts what we took to be true: $\text{det} A > 1.$
The assumption leading to the contradiction was that all eigenvalues were less than or equal to 1, so that assumption cannot be true, if the $\text{det} A > 1$. Hence, it follows, that if $\text{det} A >1$, at least one eigenvalue must be greater than 1. Which is exactly what you are to prove.