Will it be odd?

$\large \displaystyle 9^n \ mod \ 100$ , for $n \in \mathbf{N}$ , follows:

$\large \displaystyle \{ 1, 9, 81, 29, 61, 49, 41, 69, 21, 89\}$

That is:

$\large \displaystyle 9^n \ mod \ 100 = 1 \ \text{if} \ n \ mod \ 10 = 0$

$\large \displaystyle 9^n \ mod \ 100 = 9 \ \text{if} \ n \ mod \ 10 = 1$

$\large \displaystyle 9^n \ mod \ 100 = 81 \ \text{if} \ n \ mod \ 10 = 2$

$\large \displaystyle 9^n \ mod \ 100 = 29 \ \text{if} \ n \ mod \ 10 = 3$

$\large \displaystyle 9^n \ mod \ 100 = 61 \ \text{if} \ n \ mod \ 10 = 4$

$\large \displaystyle 9^n \ mod \ 100 = 49 \ \text{if} \ n \ mod \ 10 = 5$

$\large \displaystyle 9^n \ mod \ 100 = 41 \ \text{if} \ n \ mod \ 10 = 6$

$\large \displaystyle 9^n \ mod \ 100 = 69 \ \text{if} \ n \ mod \ 10 = 7$

$\large \displaystyle 9^n \ mod \ 100 = 21 \ \text{if} \ n \ mod \ 10 = 8$

$\large \displaystyle 9^n \ mod \ 100 = 91 \ \text{if} \ n \ mod \ 10 = 9$

Since the exponent is a multiple of $7$ , the last two digits of $9^{7^{5^3}}$ will be $69$ . Plus $1$ , the last two digits will be $\color{#3D99F6} \boxed{70}$ .