Here a random $n \times n$ matrix over a $Z_N$ is a matrix whose entries are uniformly random in $Z_N$. The $m$-rank of a matrix $A$ is the greatest integer $k$ such that $A$ has a $k \times k$ submatrix whose determinant is nonzero (mod m).
An article "Determinants and ranks of random matrices over Zm" gives important results on the ranks of random matrices. Define $\Pi_n(q) = (1-q) \dots (1-q^n)$. $$ \left[ \begin{array}{l} n \\ k \end{array} \right] = \frac{\Pi_n(q)}{\Pi_k(q) \Pi_{n-k}(q)} $$ In the following, $p$ is prime and $q = \frac{1}{p}$. The probability $P_{\Delta, \delta}(n,m)$ is the probability taht a random $(n + \Delta) \times n$ matrix over $Z_m$ has $m$-rank $n - \delta$. The corollary 2.2 in the article gives a formula. $A$ is a random $n \times n$ random matrix over $Z_{p^\mu}$. Then for $0 \leq i \leq p^{\mu} - 1$, $$ Pr[\det A = i \mod p^\mu] = \begin{cases} q^\mu \frac{1-q^n}{1-q} \frac{\Pi_{n+k- 1}(q)}{\Pi_{k}(q)} , & i \not = 0, gcd(i, p^\mu) = p^k\\ 1 - \frac{\Pi_{n+\mu - 1}(q)}{\Pi_{\mu-1}(q)} , & i =0 \end{cases} $$ My question is how to prove the formula in cases $i \not = 0$.