The alternating sum in row n is (-1)^(n-1)*(n - 1)!
If Y := X * (1 - X)^(z-1), then (1 - z*X)^(-1) = 1 + Sum_{n>=1} Y^n/(n-1)! * (Sum_{k=1..n} (-1)^(n-k) * z^k * T(n, k)). Note that if Y = y^(z-1) and X = x^(z-1) then y = x - x^z, dy/dx = 1 - z*x^(z-1) = 1 - z*X, and dx/dy = (1 - z*X)^(-1). Also x = y + x^z = y + y^z + z*y^(2*z-1) + ... = y * (1 + Sum_{n>=1} Y^n/(n-1)! * (1+(z-1)*n)^(-1) * (Sum_{k=1..n} (-1)^(n-k) * z^k * T(n, k))). - Michael Somos, Aug 01 2019
REFERENCES
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1998.
Formula for row n: Sum_{k = 0..n-1} T(n,k)*y^k = Product_{k = 1..n-1} (k + n*y)
E.g.f.: A(x,t) = Sum_{n >= 1} 1/(n*t)*binomial(n*t + n - 1, n)*x^n = log(B_(t+1)(x)), where B_t(x) = Sum_{n >= 0} 1/(n*t + 1)*binomial(n*t + 1, n)*x^n is Lambert's generalized binomial series - see Graham et al., Section 5.4. - Peter Bala, Nov 08 2015
T(n,m) = n^(m-1)*binomial(n-1,m-1)*Sum_{k=0..n-m} ((-1)^(n-m-k)*binomial(n+k-1,k)*stirling2(n-m+k,k)*binomial(2*n-m,n-m-k))/binomial(n-m+k,k). - Vladimir Kruchinin, Apr 05 2016