If $T_1=7^7,T_2=7^{7^{7}},T_3=7^{7^{7^{7}}}$ and so on, what will be the tens digit of $T_{1000}$? $7^4$ ends with $2301$, so $7^{4k+r}$ ends with $7^r$ digits
$7^2\equiv1 \mod(4) $, and $7^7\equiv 3 \mod(4) $
Can we use the modulo function in exponent form? I think we will use these two properties, how can I proceed further?
 A: If i correctly understand the question, we have to compute the value of $T_{1000}$ modulo $100$.
Long solution:
This is done in a few steps applying at each one the Euler indicator function $\varphi$, noting that basis $7$ is relatively prime to each of the numbers $\varphi(100)=40$, $\varphi(40)=16$, $\varphi(16)=8$ and so on, if we really need to so deeper...

We have for some odd power $N$ the relation
$$7^N=(8-1)^N\equiv(-1)^N\equiv -1\pmod8\ $$
Computations have been done in the units group of the ring $\Bbb Z/8$.

The above implies
$$
7^{7^N}
\equiv 7^{7^N\pmod {\varphi(16)}}
\equiv 7^{7^N\pmod 8}
\equiv 7^{-1}
\equiv 7\pmod{16}\ .
$$
Computations have been done in the units group of the ring $\Bbb Z/16$.

The above implies
$$
7^{7^{7^N}}
\equiv 7^{7^{7^N}\pmod {\varphi(40)}}
\equiv 7^{7^{7^N}\pmod {16}}
\equiv 7^7
\equiv 7\cdot 49^3
\equiv 7\cdot 9^3
\equiv 7\cdot 9\cdot 81
\equiv 7\cdot 9\cdot 1
\equiv 63
\equiv 23\pmod{40}\ .
$$
Computations have been done in the units group of the ring $\Bbb Z/40$.

The above implies finally
$$
7^{7^{7^{7^N}}}
\equiv 7^{7^{7^{7^N}}\pmod {\varphi(100)}}
\equiv 7^{7^{7^{7^N}}\pmod{40}}
\equiv 7^{23}
\equiv 7^3\cdot (7^4)^5
\equiv 7^3\cdot 2401^5
\equiv 7^3\cdot 1^5
\equiv 7^3
\equiv 343\equiv 63
\equiv 43\pmod{100}\ .
$$
Computations have been done in the units group of the ring $\Bbb Z/100$.
$\square$

Short solution: The above procedure is designed to work in general, this is written so for didactic purposes. In our case however, we finally need the power of an element of multiplicative order $4$ in $\Bbb Z/100$. So we need in fact in  $T_{1000}=7^{7^{\text{et caetera}}}$ only the value of
$7^{\text{et caetera}}$ modulo four. This is $-1$ (as computed even modulo $8$). So we need $7^{-1}$ in $\Bbb Z/100$. This is $43$.
A: As said in comments observe that $T_n = 7^{T_n}$.
We use induction to find tens digits of $T_n$.
Observe that $T_1 = 43 \mod 100$.
Assume that $T_n \equiv 43\mod 100$. So write $T_n = 43 + 100k$ for some $k$
So $T_{n+1} \mod 100 = 7^{T_n} \mod 100 = 7^{43} 7^{100k} \mod 100$.
Now see that $7^4 = 1 \mod 100$ so we have $7^{100k} = 1 \mod 100$ and $7^{43} \equiv 7^3 \equiv 43 \mod 100$. So we get $T_{n+1} \equiv 43 \mod 100$
So we have $T_n \equiv 43 \mod 100$ for all $n$. This means that tens digit of $T_n$ will be $43$.
A: Yes you can repeatedly apply Euler's theorem. This works because: $$a^{b^c}=a^{\left(b^c\right )}=(a^{b)\dots^b}$$
where the parenthesized exponentiation has
c parenthesization levels.  so you mod all $b$ by $\phi(n)$ then reverse the parenthesized exponentiation to unparenthesized exponentiation. Realizing equivalences in exponents mod $\phi(n)$ will reduce the next and so on.
