Find the last two digits of 7^($7^7$).

Step 1: Understand the Problem

We need to find the last two digits of $7^{7^7}$. This is equivalent to finding the value of $7^{7^7}$ modulo 100. To do this, we will use Euler's totient theorem.

Step 2: Calculate Euler's Totient Function for 100

Euler's totient function, $\phi(n)$, counts the number of positive integers up to a given integer $n$ that are relatively prime to $n$. For $n=100$, we find its prime factorization $100 = 2^2 \cdot 5^2$. Using the formula $\phi(n) = n \prod_{p|n} (1 - 1/p)$, we calculate $\phi(100)$.

Step 3: Simplify the Exponent Modulo $\phi(100)$

According to Euler's totient theorem, if $a$ and $n$ are coprime, then $a^{\phi(n)} \equiv 1 \pmod{n}$. In our case, $7$ and $100$ are coprime. So, $7^{7^7} \equiv 7^{(7^7 \pmod{\phi(100)})} \pmod{100}$. We need to calculate the exponent $7^7$ modulo $\phi(100) = 40$.

Step 4: Calculate $7^7 \pmod{40}$

We calculate the powers of $7$ modulo $40$ until we find a pattern or reach $7^7$. We find that $7^4 \equiv 1 \pmod{40}$. This simplifies the calculation for $7^7$.

Step 5: Substitute and Calculate Final Result

Now we substitute the result from the previous step back into the original expression. We need to calculate $7^{23} \pmod{100}$. We can do this by breaking down the exponent and using modular arithmetic.

Step 6: Calculate $7^{23} \pmod{100}$

We calculate powers of $7$ modulo $100$. We find that $7^4 \equiv 1 \pmod{100}$. This is a very useful property. We can then write $7^{23}$ as $(7^4)^5 \cdot 7^3$ and substitute the values.

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