Factorial problem by Ashu Dablo

Lemma: the largest integer n that can be represented by $(n-k)!$ consecutive integers is $n=(k+1)!-1$ .

I don't know a good way to explain this (or just the fact that I'm horrible at explaining things) but here it goes.

Imagine the first k numbers of $(n)!$ they are $2\cdot3\cdot4....k$

Now, We want to add another number to this while still minimizing k (explained later)

The first k+1 numbers become $1\cdot 2\cdot 3 \cdot 4...k$

Now, think of n! And $(n+1)!$ all their numbers overlap except for n+1. So $(n+1)!$ has one more number than n!

Now, remove the first k+1 digits (since we're adding one more number (this number is n+1)) and we get $n!=\dfrac{(n+1)n!}{(k+1)!}$ . (This is why we want to minimize k+1)

This implies $(k+1)!-1=n$

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Now, for the rest of our solution

Here, $n=(2^{11}+1)!-1$

It's obvious that $2^{11}$ is greater than 79. Thus $(2^{11}+1)$ contains 79,73,71 in it prime factors and is thus is $\equiv 0 \pmod{79,73,71}$

Since we are subtracting one, it will be $\equiv -1 \pmod {409457}$

Or $\equiv 409456 \pmod{409457}$

By some calculations $409456\equiv \boxed{932 \pmod{1927}}$