Hint 2 of Problem 498 Remainder of polynomial division

Problem 498

For positive integers n and m, we define two polynomials F_n(x) = xⁿ and G_m(x) = (x-1)^m.
We also define a polynomial R_n,m(x) as the remainder of the division of F_n(x) by G_m(x).
For example, R_6,3(x) = 15x² - 24x + 10.
Let C(n, m, d) be the absolute value of the coefficient of the d-th degree term of R_n,m(x).
We can verify that C(6, 3, 1) = 24 and C(100, 10, 4) = 227197811615775.
Find C(10¹³, 10¹², 10⁴) mod 999999937.

https://projecteuler.net/problem=498

Pick up where we left off earlier.
C(n,m,d) = C(n,d)* sum( C(n-d,k)*(-1)^k )    0=<k<=m-d-1

Anyone recall that C(N,K) = C(N-1,K)+C(N-1,K-1);
The equation above just makes our life a lot easier.

sum(C(N,k) * (-1)^k)  0=<k<=K
=C(N-1,K)*(-1)^K + C(N-1,K-1)*(-1)^K
+C(N-1,K-1)*(-1)^(K-1) + C(N-1,K-2)*(-1)^(K-1)
+...
+C(N-1,1)*(-1) + C(N-1,0)*(-1)
+C(N,0)*1
=C(N-1,K)*(-1)^K

That is to say, 
C(n,m,d) = C(n,d)* sum( C(n-d,k)*(-1)^k )    0=<k<=m-d-1
=> C(n,m,d) = C(n,d)* C(n-d-1,m-d-1)*(-1)^(m-d-1)  

In this case,
C(10^13,10^12,10^4)
= C(10^13 , 10^4) * C(10^13-10^4-1 , 10^12-10^4-1) * (-1)

Since it is an absolute value of d-th degree, we can get rid of the (-1).
C(10^13,10^12,10^4) = C(10^13 , 10^4) * C(10^13-10^4-1 , 10^12-10^4-1)

No Sum, a lot better, right?
But still, things can get even better!
TBC in Hint3!

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The opinion above is original and unproven work. Please correct me if I made a mistake.
Please tell me if you have reached the correct answer using the methods above.
I am just too lazy to calculate that. :)