Last two digits in nested powers of 7

Exercise:

What are the last two digits of ${{7}^{{{7}^{{{7}^{{{7}^{{{7}^{{{7}^{7}}}}}}}}}}}}$

Solution: Let $a={{7}^{7}}$ be last index on the given number .

We know that ${{7}^{2}}=49\equiv -51\left( \bmod 100 \right)$$\Rightarrow {{7}^{3}}=\left( -51 \right)\times 7=-357+300=-57\equiv 43\left( \bmod 100 \right)$

So the given number will become $b={{7}^{{{7}^{{{7}^{{{7}^{{{7}^{43}}}}}}}}}}$ Now take $w={{7}^{43}}$ and let try to find the last 2 digits as did in the above ${{7}^{14}}={{7}^{7}}\times {{7}^{7}}=\left( 43 \right)\left( 43 \right)\equiv 49\left( \bmod 100 \right)$

hence ${{7}^{42}}={{\left( {{7}^{14}} \right)}^{3}}={{\left( 49 \right)}^{3}}\equiv 49\left( \bmod 100 \right)$ thus ${{7}^{43}}={{7}^{42}}\times 7=7\times \left( 49 \right)\equiv 43\left( \bmod 100 \right)$

thus the give number will transform to this form $b={{7}^{{{7}^{{{7}^{43}}}}}}$ which is same as $w$

thus ${{7}^{43}}\equiv 43\left( \bmod 100 \right)$ so the last 2 digits are $43$