Find the last digit of `32^(32^(32))`

Answer

To find the last digit of \( 32^{32^{32}} \), we can follow these steps:

### Step 1: Identify the base and its last digit
The base is \( 32 \). The last digit of \( 32 \) is \( 2 \).

### Step 2: Simplify the expression
We need to find the last digit of \( 2^{32^{32}} \).

### Step 3: Determine the cycle of last digits for powers of 2
The last digits of the powers of \( 2 \) follow a repeating cycle:
- \( 2^1 = 2 \) (last digit is \( 2 \))
- \( 2^2 = 4 \) (last digit is \( 4 \))
- \( 2^3 = 8 \) (last digit is \( 8 \))
- \( 2^4 = 16 \) (last digit is \( 6 \))
- \( 2^5 = 32 \) (last digit is \( 2 \)) and the cycle repeats every 4 terms: \( 2, 4, 8, 6 \).

### Step 4: Find the exponent modulo 4
To determine which last digit we will use, we need to find \( 32^{32} \mod 4 \):
- Since \( 32 \equiv 0 \mod 4 \), any power of \( 32 \) will also be \( 0 \mod 4 \).

### Step 5: Determine the position in the cycle
Since \( 32^{32} \equiv 0 \mod 4 \), we look at the last digit corresponding to \( 2^0 \) in the cycle:
- The last digit for \( 2^0 \) is \( 6 \) (since \( 2^4 \) corresponds to \( 6 \)).

### Conclusion
Thus, the last digit of \( 32^{32^{32}} \) is \( 6 \).

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