We are tasked with finding the remainder when the expression

$\left[ (1+10!) \cdot \left(1+(10!)^2\right) \cdot \dots \cdot \left(1+(10!)^{100}\right) \right]^{100}$

is divided by $10!$.

Step 1: Simplifying the expression mod $10!$

Let $N = 10! = 3,628,800$. We are interested in calculating the expression modulo $10!$.

We can rewrite each term in the product as $1 + (10!)^k$ for $k = 1, 2, \dots, 100$. Notice that for any $k \geq 1$, $(10!)^k$ is divisible by $10!$. Therefore, we have:

$1 + (10!)^k \equiv 1 \pmod{10!}$

This means that each term in the product $\left(1 + (10!)^k\right)$ is congruent to 1 modulo $10!$.

Step 2: Entire product mod $10!$

Since each term in the product is congruent to 1 modulo $10!$, the entire product is:

$\left( (1+10!) \cdot (1+(10!)^2) \cdot \dots \cdot (1+(10!)^{100}) \right) \equiv 1 \cdot 1 \cdot \dots \cdot 1 = 1 \pmod{10!}$

Step 3: Raising to the power of 100

Now, we raise the entire product to the power of 100:

$\left( \left[ (1+10!) \cdot (1+(10!)^2) \cdot \dots \cdot (1+(10!)^{100}) \right]^{100} \right) \equiv 1^{100} = 1 \pmod{10!}$

Step 4: Final result

Therefore, the remainder when the expression is divided by $10!$ is:

$\boxed{1}$

Follow-up questions:

1. How would the result change if $10!$ were replaced by $9!$?
2. What is the significance of modulo in solving this type of problem?
3. Can the same approach be used for powers greater than 100 in this type of expression?
4. How would the result be affected if instead of powers of $10!$, we had powers of another factorial, like $8!$?
5. Why does $(10!)^k$ being divisible by $10!$ simplify the problem?

Tip:

When working with factorials and large powers, recognizing patterns of divisibility can drastically simplify complex problems.