The cells of a $2021\times 2021$ table are filled with numbers using the following rule. The bottom left cell, which we label with coordinate $(1, 1)$, contains the number $0$. For every other cell $C$, we consider a route from $(1, 1)$ to $C$, where at each step we can only go one cell to the right or one cell up (not diagonally). If we take the number of steps in the route and add the numbers from the cells along the route, we obtain the number in cell $C$. For example, the cell with coordinate $(2, 1)$ contains $1 = 1 + 0$, the cell with coordinate $(3, 1)$ contains $3 = 2 + 0 + 1$, and the cell with coordinate $(3, 2)$ contains $7 = 3 + 0 + 1 + 3$. What is the last digit of the number in the cell $(2021, 2021)$?

1. **Understanding the Problem:**
   - We have a $2021 \times 2021$ table.
   - The bottom-left cell $(1, 1)$ contains the number $0$.
   - For any other cell $(x, y)$, the number in the cell is determined by the sum of the steps taken to reach that cell from $(1, 1)$ plus the sum of the numbers in the cells along the route.

2. **Initial Observations:**
   - The cell $(2, 1)$ contains $1 = 1 + 0$.
   - The cell $(3, 1)$ contains $3 = 2 + 0 + 1$.
   - The cell $(3, 2)$ contains $7 = 3 + 0 + 1 + 3$.

3. **General Formula:**
   - Let $C_n$ be the value in cell $(x, y)$ where $n = x + y - 2$.
   - We claim that $C_n = 2^n - 1$ and prove this by induction.

4. **Base Case:**
   - For $n = 0$, $C_0 = 2^0 - 1 = 0$, which is true since the cell $(1, 1)$ contains $0$.

5. **Inductive Step:**
   - Assume the proposition holds for $C_k = 2^k - 1$.
   - We need to prove $C_{k+1} = 2^{k+1} - 1$.
   - By definition:
     \[
     C_{k+1} = (k+1) + \sum_{0 \le i < k+1} C_i
     \]
   - Using the inductive hypothesis:
     \[
     C_{k+1} = (k+1) + \sum_{0 \le i < k} C_i + C_k
     \]
   - Since $\sum_{0 \le i < k} C_i = 2^k - 1$ and $C_k = 2^k - 1$:
     \[
     C_{k+1} = (k+1) + (2^k - 1) + (2^k - 1)
     \]
     \[
     C_{k+1} = (k+1) + 2 \cdot (2^k - 1)
     \]
     \[
     C_{k+1} = (k+1) + 2^{k+1} - 2
     \]
     \[
     C_{k+1} = k + 1 + 2^{k+1} - 2
     \]
     \[
     C_{k+1} = 2^{k+1} - 1
     \]
   - This completes the induction proof.

6. **Conclusion:**
   - The number in the cell $(2021, 2021)$ is $C_{4040} = 2^{4040} - 1$.
   - We need to find the last digit of $2^{4040} - 1$.
   - The last digit of powers of $2$ cycles every 4: $2, 4, 8, 6$.
   - Since $4040 \mod 4 = 0$, the last digit of $2^{4040}$ is $6$.
   - Therefore, the last digit of $2^{4040} - 1$ is $6 - 1 = 5$.

The final answer is $\boxed{5}$.