Lets make the expansion of the next sum around $h=1$ $$ \left ( \prod_{n=1}^{N} \ \sum_{l_n=0}^{N_f} \ h^{2 l_n} \ \binom{N_f}{l_n} \right ) \ \underset{1\le i,j\le N}{\det} \binom{N_f}{l_i-i+j} =|h^{2 l_n} =1+2 l_n \log h + ... | $$ $$ = \underset{1\le i,j\le N}{\det} \binom{N_f^2}{N_f-i+j} + \log h \sum_{k=1}^N \underset{1\le i,j\le N}{\det} (N_f+ k-j) \binom{N_f^2}{N_f -i+j} + $$ $$ + (\log h)^2 \left \{ \sum_{k=1}^N \underset{1\le i,j\le N}{\det} \frac{(N_f+ k-j)(N_f^2+(1-N_f)(k-j))}{2 N_f -1} \binom{N_f^2}{N_f-i+j} \right. + $$ $$ \left. +\sum_{k, l=1}^N \underset{1\le i,j\le N}{\det} (N_f+ k-j) (N_f+ l-j) \binom{N_f^2}{N_f-i+j} \right \} + $$ $$ + (\log h)^3\left \{\sum_{k=1}^N \underset{1\le i,j\le N}{\det}\!\!\!\! \frac{ (N_f+k-j)^2 \left(N_f(N_f+1)+ (2- N_f)(k-j)\right)}{3(2 N_f-1)} \!\!\binom{N_f^2}{N_f-i+j}+ ... \right \}+ O( (\log h)^4) . $$ Using $$ \sum_{s=0}^{N_f} \ \binom{N_f}{s} \binom{N_f}{s-k +l} = \binom{2 N_f}{N_f-k+l} $$ we came to \begin{eqnarray} \underset{1\le k,l \le N}{\det} \ \binom{2 N_f}{N_f-k+l} = \frac{ G[N+ 2 N_f+1] G[N+1] G[N_f+1]^2 } { G[2 N_f+1] G[N+ N_f+1]^2 }\ = Z_0 \ . \end{eqnarray} First correction is simple $$ \sum_{k=1}^N \underset{1\le i,j\le N}{\det} (N_f+ k-j) \binom{N_f^2}{N_f-i+j} = N_f N \, Z_0 $$ because of $$ \sum_{k=1}^N \underset{1\le i,j\le N}{\det} ( k-j) \binom{N_f^2}{N_f-i+j} = \epsilon^{j_1 .. j_N} [(1- j_1)+... +(N- j_N)] $$ $$ \times\binom{N_f^2}{N_f-1+j_1}\cdot ... \cdot \binom{N_f^2}{N_f-N+j_N} = 0 \ . $$ But with next terms we get in trouble (looking at $- 2k\cdot j$ terms) $$ \sum_{k=1}^N \underset{1\le i,j\le N}{\det} ( k-j)^2 \binom{N_f^2}{N_f-i+j} = \epsilon^{j_1 .. j_N} \left \{ [(1+ j_1^2)+... +(N^2+ j_N^2)] + \right. $$ $$ \left. - 2( j_1+ 2 j_2+ ...+ N j_N) \right \} \binom{N_f^2}{N_f-1+j_1}\cdot ... \cdot \binom{N_f^2}{N_f-N+j_N}= 2 \frac{N(N+1)(2 N+1)}{6} Z_0 + ??? $$ How is to get determinants of this type?