Exponentiation of Pascal's Triangle(in matrix form) I want to find a pattern in subsequent exponentiations of the pascal triangle shown in the form below
Matrix P[K+1][K+1]:
$$
        \begin{matrix}
        \binom{0}{0} & 0 & 0 & 0\cdots &0\\
        \binom{1}{0} & \binom{1}{1} & 0 & 0\cdots &0 \\
        \binom{2}{0} & \binom{2}{1} & \binom{2}{2} & 0\cdots  &0 \\
        \binom{3}{0} & \binom{3}{1} & \binom{3}{2} & \binom{3}{3}\cdots &0 \\
        \vdots & \vdots  &\vdots  &\vdots\ddots& \vdots\\
        \binom{k}{0} & \binom{k}{1}& \binom{k}{2} & \binom{k}{3}\cdots &\binom{k}{k} \\
\end{matrix}
$$
My motive is to compute $P^N$ in $O(KlogN)$ complexity ,which is definately not possible for any ordinary matrix(which would take $O(K^3logN)$).
But I am pretty much sure it is possible for this special matrix as I have done it in $O(K^2logN)$ but the problem(here),which i am solving,suggests
it to be done in $O(KlogN)$,Definately there is a secret which I want to find.
Thanks in Advanced.

You can use a binomial expansion and rearrangement to show 
$$(1+N)^{i-j}{i\choose j}=\sum_{n=0}^{i-j} N^{i-j-n}{i-j \choose n} {i\choose j} =\sum_{m=j}^i N^{i-m}{i\choose m} {m\choose j} $$
and then induction to show 
$$P^N=\left(\begin{matrix}
        N^{0-0}\binom{0}{0} & 0 & 0 & 0&\cdots &0\\
        N^{1-0}\binom{1}{0} & N^{1-1}\binom{1}{1} & 0 & 0&\cdots &0 \\
        N^{2-0}\binom{2}{0} & N^{2-1}\binom{2}{1} & N^{2-2}\binom{2}{2} & 0&\cdots  &0 \\
        N^{3-0}\binom{3}{0} & N^{3-1}\binom{3}{1} & N^{3-2}\binom{3}{2} & N^{3-3}\binom{3}{3}&\cdots &0 \\
        \vdots & \vdots  &\vdots  &\vdots&\ddots& \vdots\\
        N^{k-0}\binom{k}{0} & N^{k-1}\binom{k}{1}& N^{k-2}\binom{k}{2} & N^{k-3}\binom{k}{3}&\cdots &N^{k-k}\binom{k}{k} \\
\end{matrix}\right)$$
which will give you a fast calculation of any power of $P$ without having to find the intermediate powers.  
