For any $N$, let $f(N)$ be the last five digits before the trailing zeroes in $N!$.
For example,

• $9! = 362880$ so $f(9)=36288$
• $10! = 3628800$ so $f(10)=36288$
• $20! = 2432902008176640000$ so $f(20)=17664$

Find $f(1\,000\,000\,000\,000)$.