Computing $75^{75^{75}}$ modulo $32$

Question

How to compute $75^{75^{75}}$ modulo $32$?

I tried:

It's $\mathrm{gcd}(75,32)=1$, so with Euler's phi function I get

$\varphi(32)=32(1-\frac{1}{2})=16$.

Then it's $75^{16}\equiv1 \ (\mathrm{mod} \ 32)$.

So I computed $75^{75}$ modulo $16$:

It's $\varphi(16)=16(1-\frac{1}{2})=8$

So $75^{75}\equiv 1 \ (\mathrm{mod} \ 8)$ and $75^{75^{75}} \equiv 1 \ (\mathrm{mod} \ 8)$

Then it's $75^{75^{75}}\equiv 75^1 \equiv 75 \ (\mathrm{mod} \ 32)$.

I'm not sure if this is correct. Is there something wrong in this calculation or can it be done like this?

Answer 1

$$75^{75 } \equiv 75^{72} 75^3 \equiv (75^{8})^{9}11^3 \equiv 3 \mod 16$$

so we can say $$ {75^{75}}^{75} \equiv 75^3 \equiv 11^3 \equiv 19 \mod 32$$