Question 852815
From this page, <a href="http://www.math.lsa.umich.edu/~hochster/419/det.html">http://www.math.lsa.umich.edu/~hochster/419/det.html</a>, it states that 


<blockquote><font color="blue">Fact 7. The determinant of a lower triangular matrix (or an upper triangular matrix) is the product of the diagonal entries. In particular, the determinant of a diagonal matrix is the product of the diagonal entries.</font></blockquote> 

Since matrix A=([1,0,0,0,0],[a,-2,0,0,0],[a^2,b,-3,0,0],[a^3,b^2,c,-4,0],[a^4,b^3,c^2,d,-5]), which really looks like this


[1,0,0,0,0]
[a,-2,0,0,0]
[a^2,b,-3,0,0]
[a^3,b^2,c,-4,0]
[a^4,b^3,c^2,d,-5]


is <a href="http://mathworld.wolfram.com/LowerTriangularMatrix.html">lower triangular</a>, this means that we just multiply the diagonal entries, which are: 1, -2, -3, -4, -5


So multiplying them gives us: 1*(-2)*(-3)*(-4)*(-5) = 120


which is equivalent to 5! since 5! = 5*4*3*2*1 = 120


The answer is <font size=4 color="red">a)|A|=5!</font>