The digit at the tens place in the sum of the expression :
`(1!) + (2 !)^(2) + (3 !)^(3) + (4 !)^4 + (5 !)^(5) + (111!)^(111)` is :

Options

A. 0
B. 1
C. 8
D. 9

Answer

To find the digit at the tens place in the sum of the expression:

\[ (1!) + (2!)^2 + (3!)^3 + (4!)^4 + (5!)^5 + (111!)^{111} \]

we will evaluate each term step by step.

### Step 1: Calculate \(1!\)
\[ 1! = 1 \]

### Step 2: Calculate \((2!)^2\)
\[ 2! = 2 \quad \Rightarrow \quad (2!)^2 = 2^2 = 4 \]

### Step 3: Calculate \((3!)^3\)
\[ 3! = 6 \quad \Rightarrow \quad (3!)^3 = 6^3 = 216 \]

### Step 4: Calculate \((4!)^4\)
\[ 4! = 24 \quad \Rightarrow \quad (4!)^4 = 24^4 \]
To find \(24^4\), we can calculate it step by step:
\[ 24^2 = 576 \quad \Rightarrow \quad 24^4 = 576^2 \]
Calculating \(576^2\):
\[ 576^2 = (500 + 76)^2 = 500^2 + 2 \times 500 \times 76 + 76^2 \]
\[ = 250000 + 76000 + 5776 = 326776 \]

### Step 5: Calculate \((5!)^5\)
\[ 5! = 120 \quad \Rightarrow \quad (5!)^5 = 120^5 \]
Since \(120\) ends with \(0\), any power of \(120\) will also end with \(00\). Thus, the last two digits of \((5!)^5\) are \(00\).

### Step 6: Calculate \((111!)^{111}\)
For \(n \geq 5\), \(n!\) will end with \(0\) because \(5!\) and higher factorials contain both \(2\) and \(5\) as factors. Therefore, \((111!)^{111}\) will also end with \(00\).

### Step 7: Sum the contributions
Now we sum the contributions from each term:
- \(1! = 1\)
- \((2!)^2 = 4\)
- \((3!)^3 = 216\)
- \((4!)^4 = 326776\)
- \((5!)^5 = 00\)
- \((111!)^{111} = 00\)

Adding these values:
\[ 1 + 4 + 216 + 326776 + 0 + 0 = 326997 \]

### Step 8: Identify the digit at the tens place
The number \(326997\) has the tens digit as \(9\).

### Final Answer
Thus, the digit at the tens place in the sum is:

\[ \boxed{9} \]