1. Generalization and Induction Hypothesis:
We generalize the problem by replacing $100$ with $n \geq 3$. We aim to prove the statement by induction on $n$, starting with the base case $n=4$.

2. Induction Step:
Assume the statement is true for some $n-1$, i.e., there exist at least $(n-1)!$ ways to write $(n-1)!$ as a sum of elements from the set $\{1!, 2!, \ldots, (n-1)!\}$.

3. **Expanding $n!$ with $i$ $(n-1)!$'s:**
For any nonnegative integer $i \leq n-1$, consider the number of ways to expand $n!$ as a sum of elements from $\{1!, 2!, \ldots, n!\}$ with exactly $i$ $(n-1)!$'s. After removing $i$ $(n-1)!$'s, the remaining sum is:
$$n! - i \cdot (n-1)! = (n-i)(n-1)!$$
By the induction hypothesis, there are at least $(n-1)!$ ways to expand $(n-i)(n-1)!$ using elements from $\{1!, 2!, \ldots, (n-1)!\}$.

4. Counting the Total Number of Ways:
Since we have $n$ cases (one for each $i$ from $0$ to $n-1$), and each case has more than $(n-1)!$ ways, the total number of ways to write $n!$ as a sum of elements from $\{1!, 2!, \ldots, n!\}$ is:
$$n \cdot (n-1)! > n!$$
This completes the induction step.

5. **Base Case $n=4$:**
We manually verify the base case for $n=4$:
- If there are no $3!$, then we have $4!$ ways.
- If there is one $3!$, then we have $3!$ ways.
- If there are two $3!$'s, then we have $2!$ ways.
- If there are three $3!$'s, then we have $1!$ ways.
- If there are four $3!$'s, then we have $0!$ ways.

Summing these, we get:
$$4! + 3! + 2! + 1! + 0! = 24 + 6 + 2 + 1 + 1 = 34 > 4!$$
Thus, the base case holds.

By induction, we have shown that there exist at least $100!$ ways to write $100!$ as a sum of elements from the set $\{1!, 2!, \ldots, 99!\}$.

$\blacksquare$