For integer ` n gt 1`, the digit at unit's place in the number
` sum_(r=0)^(100) r! + 2^(2^(n))` I

Answer

To find the digit at the unit's place in the expression \( \sum_{r=0}^{100} r! + 2^{2^n} \) for integer \( n > 1 \), we will break down the problem into manageable steps.

### Step 1: Calculate the sum \( \sum_{r=0}^{100} r! \)

The sum of factorials from \( 0! \) to \( 100! \) is given by:

\[ 0! + 1! + 2! + 3! + 4! + 5! + \ldots + 100! \]

Calculating the first few factorials:

- \( 0! = 1 \)
- \( 1! = 1 \)
- \( 2! = 2 \)
- \( 3! = 6 \)
- \( 4! = 24 \)
- \( 5! = 120 \)

Now, let's add these values:

\[ 0! + 1! + 2! + 3! + 4! + 5! = 1 + 1 + 2 + 6 + 24 + 120 = 154 \]

### Step 2: Analyze the unit's place of the sum

For \( r \geq 5 \), \( r! \) will end with a zero because \( 5! \) and higher factorials contain at least one factor of \( 10 \) (from \( 2 \times 5 \)). Therefore, the unit's place of the sum \( \sum_{r=0}^{100} r! \) will be determined only by the first five terms:

\[ 154 \quad \text{(unit's place is 4)} \]

### Step 3: Calculate \( 2^{2^n} \)

Next, we need to find the unit's place of \( 2^{2^n} \).

For \( n > 1 \), we can evaluate \( 2^{2^n} \) for the first few values of \( n \):

- For \( n = 2 \): \( 2^{2^2} = 2^4 = 16 \) (unit's place is 6)
- For \( n = 3 \): \( 2^{2^3} = 2^8 = 256 \) (unit's place is 6)
- For \( n = 4 \): \( 2^{2^4} = 2^{16} = 65536 \) (unit's place is 6)

We can observe that for \( n \geq 2 \), the unit's place of \( 2^{2^n} \) is consistently 6.

### Step 4: Combine the results

Now we combine the results from Steps 2 and 3:

\[ \text{Unit's place of } \sum_{r=0}^{100} r! + 2^{2^n} = 4 + 6 = 10 \]

Thus, the unit's place of the entire expression is:

\[ \text{Unit's place is } 0 \]

### Final Answer

The digit at the unit's place in the number \( \sum_{r=0}^{100} r! + 2^{2^n} \) for integer \( n > 1 \) is **0**.