<Question>
What's the largest eight-digit base 2 integer?  Express your answer in base 10.
</Question>

<think>
<Plan_1>
Okay, so I need to find the largest eight-digit base 2 integer and then express that number in base 10. Hmm, let me start by recalling what base 2 means.
Base 2, or binary, uses digits 0 and 1, and each digit represents a power of 2. The rightmost digit is 2^0, then 2^1, 2^2, and so on. So, for an eight-digit binary number, the leftmost digit is the 2^7 place.

Wait, if it's eight digits, then the largest number would be when all the digits are 1.
Let me check that. For example, in a four-digit binary number, the largest is 1111, which is 15 in decimal. So, following that logic, an eight-digit binary number would be 11111111.

Let me confirm that. Each digit from right to left is 1, so the places are 2^7, 2^6, ...., 2^0. So adding all those up.
The sum would be 2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^1 + 2^0.

I remember that the sum of a geometric series where each term is double the previous one is 2^(n) - 1, where n is the number of terms. Here, there are 8 terms, so the sum should be 2^8 - 1.
Let me compute that: 2^8 is 256, so 256 - 1 is 255. So, 11111111 in binary is 255 in decimal.

Is there a larger eight-digit binary number? Well, if all digits are 1, that's the maximum. If any digit is 0, then the number would be smaller.
For example, 11111110 is 254, which is one less. So yes, 11111111 is indeed the largest.

Wait, but let me make sure I'm not confusing the number of digits with the exponent. An eight-digit binary number ranges from 00000000 (0) to 11111111 (255). So the maximum is 255.

Therefore, the answer should be 255 in base 10.

Let me double-check by converting 11111111 from binary to decimal step by step.

<prompt_1.1> Double-check the conversion of the binary number 11111111 to decimal by calculating each bit's value. </prompt_1.1>

Alternatively, using the formula for the sum of a geometric series: a*(r^n -1)/(r-1). Here, a=1, r=2, n=8.
So, (2^8 -1)/(2-1) = 255/1 = 255.

Another way to think: each time you add a digit, you double the previous total and add 1.

<prompt_1.2> Verify the sum of the binary number 11111111 using an iterative method. </prompt_1.2>

So, all methods confirm that the largest eight-digit binary number is 255 in decimal.

Just to ensure there's no mistake in the number of digits: 11111111 has eight 1's.

<prompt_1.3> Confirm that the binary number 11111111 has eight digits and verify the final answer. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
The rightmost digit is 1, which is 2^0 = 1. Next is 2^1 = 2, then 4, 8, 16, 32, 64, 128. Adding all those: 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1. Let's add them step by step:

128 + 64 = 192

192 + 32 = 224

224 + 16 = 240

240 + 8 = 248

248 + 4 = 252

252 + 2 = 254

254 + 1 = 255.
</execution_1.1>

<execution_1.2>
For example, starting from the left:

First digit (leftmost) is 1 * 128 (2^7). Then adding the next 1 * 64 (2^6) gives 192. Then 1 * 32 (2^5) makes 224. Then 16 (2^4) brings it to 240. Next, 8 (2^3) gives 248. Then 4 (2^2) is 252. Adding 2 (2^1) gives 254. Finally, adding 1 (2^0) gives 255.
</execution_1.2>

<execution_1.3>
So starting from the left, each digit is 2^7 down to 2^0. So yes, eight digits. Therefore, converting that gives 255. 

I think that's thorough. I considered different methods of conversion, checked the formula for the sum of a geometric series, verified with an example with fewer digits, and confirmed the number of digits in the binary number. All lead to the same conclusion.
</execution_1.3>
</Execution_1>

<Final_answer>
Therefore, the largest eight-digit base 2 integer is $\boxed{255}$ in base 10.
</Final_answer>
</think>