Successive squaring with big exponent I would like to calculate $ 2^{5^{77}}\bmod 113 $.
I solved similar problems but with smaller exponent e.g. $ 2^{185}\bmod 113 $ with successive squaring method. Is there any way to use this method to solve it?
Rather than successive squaring, try successive raising to the $5^{th}$ power:

$2^{5^0}\equiv2\bmod 113$

$2^{5^1}\equiv32\bmod113$

$2^{5^2}\equiv32^5\equiv99\bmod113$

$2^{5^3}\equiv99^5\equiv56\bmod113$

$2^{5^4}\equiv56^5\equiv60\bmod113$

$2^{5^5}\equiv60^5\equiv105\bmod113$

$2^{5^6}\equiv105^5\equiv\color{red}2\bmod113$.

Can you take it from here?   $\;$  Note that $77\equiv5\bmod6$.
You can reduce the exponent by Lil Fermat.  You need $5^{77}\bmod{112}$.

Now by Euler's theorem, since $\varphi(112)=48$, we $5^{48}\cong1\bmod{112}$.  So $5^{77}\cong5^{29}\bmod{112}$ and we can proceed with successive squaring.

$5^3\cong13\bmod{112}\implies5^6\cong57\bmod{112}\implies5^{12}\cong1\bmod{112}\implies5^{24}\cong1\bmod{112}\implies5^{29}\cong5^5\bmod{112}\implies5^{29}\cong-11\bmod{112}\implies5^{29}\cong101\bmod{112}$.

So $2^{5^{77}}\cong2^{101}\bmod{113}$.

Now we do more successive squaring and the like: $2^{10}\cong7\bmod{113}\implies2^{20}\cong49\bmod{113}\implies2^{30}\cong{4}\bmod{113}\implies2^{60}\cong16\bmod{113}\implies2^{90}\cong64\bmod{113}\implies2^{100}\cong-4\bmod{113}\implies2^{101}\cong-8\bmod{113}$.

Thus $2^{5^{77}}\cong105\bmod{113}$.