First of all, the matrix isn't quite right, but I didn't notice that the first time I looked at it. That may be your only issue, but I'll be explicit about everything in case something else is confusing you.
Does it make sense that you can view polynomials in a vector space? Just like in a vector space, the coefficients of any single term in the polynomial do not effect the other terms. If you have 5x^3, the 5 only effects the "x^3 direction". When you add up all of the terms, the number of x^3 added in will be 5 no matter how many times x^2 you also add in. Yes, when you evaluate there may be different values of x such that the polynomial equals the same thing, but we are not dealing with final value when we differentiate. The map for differentiation is,
<br />
D: \mathbb{R}[x]\rightarrow\mathbb{R}[x]<br /> Hopefully that part makes sense. If not, make sure that you are comfortable with the idea that polynomials fit the requirements to be a vector space.
When you differentiate a single power of x, the exponent is multiplied by the coefficient and then that value (coeff x exponent) is moved into the position as coefficient for one power down of x. In calc you think about this as subtracting one from the exponent, but here you are being asked to imagine that all (up to N) of the possible values of x^n exist, but may be multiplied by zero the same way that a vector point down the x-axis doesn't mean that y\mbox{ and } z disappear, they just have zeros for coefficients.
So our N+1 dimensional vector [c_0,c_1,c_2,\ldots,c_N]^T of the coefficients of the polynomial c_0x^0+c_1x^1+c_2x^2+\ldots+c_Nx^N can be written as a column vector with c_0 at the top and c_N at the bottom. When you write the matrix for D, it puts c_1 in the c_0 place and gets rid of c_0. That is just the normal rule from calculus. Then c_n*n\rightarrow c_{n-1} follows in general so the matrix will have 1,2,3,4... just off the diagonal
<br />
\begin{array}{cccccc}<br />
0&1&0&0&\ldots&0\\<br />
0&0&2&0&\ldots&0\\<br />
0&0&0&3&\ldots&0\\<br />
\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\<br />
0&0&0&0&\ldots&N\\<br />
0&0&0&0&\ldots&0\\<br />
\end{array}<br />
When you square this matrix, you will get<br />
\begin{array}{cccccc}<br />
0&0&2&0&\ldots&0\\<br />
0&0&0&2*3&\ldots&0\\<br />
\ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\<br />
0&0&0&0&\ldots&N(N-1)\\<br />
0&0&0&0&\ldots&0\\<br />
0&0&0&0&\ldots&0\\<br />
\end{array}<br />
Which is the same as taking the derivative twice. If you are not convinced, try an easy polynomial like 1+x+x^2+x^3+x^4+x^5 using your normal calc rules and then do it with a matrix. Then you can put the c_n in as coefficients to finally remove all doubt.
