How to show that det(A)≤1? 


Let $A = (a_{ij})_n$ where $a_{ij} \ge 0$ for $i,j=1,2,\ldots,n$ and $\sum_{j=1}^n a_{ij} \le 1$ for $i = 1,2,\ldots,n$.
  Show that $|\det(A)| \le 1$.

Should I use the definition of matrix:
$$\det(A)=\sum \textrm{sgn}(\sigma)a_{1,\sigma(1)}a_{2,\sigma(2)}...a_{n,\sigma(n)}$$
I don't understand what is $a_{i,\sigma(i)}$ ? Where $i=1,2,
\ldots,n$. Or is there another way to solve it?

Just take modulus the determinant expression you wrote. Note each term on the right under sum can be bounded by using G.M. $\le$ A.M.
