Compute $\left\lceil\displaystyle\sum_{k=2018}^{\infty}\frac{2019!-2018!}{k!}\right\rceil$. (The notation $\left\lceil x\right\rceil$ denotes the least integer $n$ such that $n\geq x$.)

[i]Proposed by Tristan Shin[/i]
1. We start with the given expression:
   \[
   \left\lceil \sum_{k=2018}^{\infty} \frac{2019! - 2018!}{k!} \right\rceil
   \]
   We can factor out the common term \(2018!\) from the numerator:
   \[
   2019! - 2018! = 2018! \cdot (2019 - 1) = 2018! \cdot 2018
   \]
   Thus, the expression becomes:
   \[
   \left\lceil \sum_{k=2018}^{\infty} \frac{2018 \cdot 2018!}{k!} \right\rceil
   \]

2. We can simplify the fraction inside the summation:
   \[
   \frac{2018 \cdot 2018!}{k!} = 2018 \cdot \frac{2018!}{k!}
   \]
   Therefore, the expression is:
   \[
   \left\lceil 2018 \sum_{k=2018}^{\infty} \frac{2018!}{k!} \right\rceil
   \]

3. We now consider the series \(\sum_{k=2018}^{\infty} \frac{2018!}{k!}\). Notice that:
   \[
   \frac{2018!}{2018!} = 1
   \]
   \[
   \frac{2018!}{2019!} = \frac{1}{2019}
   \]
   \[
   \frac{2018!}{2020!} = \frac{1}{2019 \cdot 2020}
   \]
   and so on. The series can be written as:
   \[
   \sum_{k=2018}^{\infty} \frac{2018!}{k!} = 1 + \frac{1}{2019} + \frac{1}{2019 \cdot 2020} + \frac{1}{2019 \cdot 2020 \cdot 2021} + \cdots
   \]

4. The series \(\sum_{k=2018}^{\infty} \frac{2018!}{k!}\) converges to a value slightly greater than 1. This is because the terms \(\frac{1}{2019}, \frac{1}{2019 \cdot 2020}, \ldots\) are very small.

5. Therefore, we can approximate the sum:
   \[
   \sum_{k=2018}^{\infty} \frac{2018!}{k!} \approx 1 + \epsilon
   \]
   where \(\epsilon\) is a very small positive number.

6. Multiplying by 2018, we get:
   \[
   2018 \sum_{k=2018}^{\infty} \frac{2018!}{k!} \approx 2018 \cdot (1 + \epsilon) = 2018 + 2018\epsilon
   \]
   Since \(\epsilon\) is very small, \(2018\epsilon\) is also very small, and the expression is very close to 2018.

7. Taking the ceiling of this value:
   \[
   \left\lceil 2018 + 2018\epsilon \right\rceil = 2019
   \]

The final answer is \(\boxed{2019}\).