Finding determinant of an $n\times n$ matrix whose $(i, j)$th element is $i^j$ I tried using induction and partition matrix.
Let $A$ be an $(n-1) \times (n-1)$ matrix of the same type. 
By my method, if I show that $(n . n^2\ldots n^{n-1} ) A^{-1} (1 2^n\ldots(n-1)^n)^T = n^n -n!$ , then I am done. 
But it seems rather tough to show that. Is there an easier method? 
 A: Let $A$ be a $n\times n$ matrix (with $n\geq 2$) such that $A_{i,j}=i^j$. Writing
$$ A (c_1,c_2,\ldots,c_n)^T = (y_1,y_2,\ldots,y_n)^T $$
is the same as stating that the polynomial $p(x)=c_1 x+c_2 x^2+\ldots+c_n x^n$ attains the values $y_1,y_2,\ldots,y_n$ at $1,2,\ldots,n$, or that the polynomial $q(x)=\frac{p(x)}{x}=c_1+\ldots+c_n x^{n-1}$ attains the values $\frac{y_1}{1},\frac{y_2}{2},\ldots,\frac{y_n}{n}$ at $1,2,\ldots,n$. The problem of finding $(c_1,c_2,\ldots,c_n)^T$, given $(y_1,y_2,\ldots,y_n)^T$, has a unique solution provided by Lagrange interpolation. For instance, assuming $(y_1,y_2,\ldots,y_n)^T=(0,0,\ldots,1)^T$ we have
$$ q(x) = \frac{1}{n!}(x-1)(x-2)\cdot\ldots\cdot(x-n+1) $$
so $c_n=\frac{1}{n!}$. On the other hand Cramer's rule gives
$$ c_n = \frac{\det A^{(n)}}{\det A}, $$
with $A^{(n)}$ being the $A$-matrix without the last row and column, i.e. the $A$-matrix in dimension $(n-1)$. By induction it follows that
$$ \det A = n!\cdot(n-1)!\cdot\ldots\cdot 1! = \prod_{k=1}^{n} k^{n+1-k}.$$
