Calculate it

((2)^7)^6) * 2^6/ 97= (128)^6 * 64 / 97 = (31)^6 (-33)/97= (961)^3 * (-33) / 97= (-9)^3 (-33)/ 97= 24057/97= rem. 1

Technically, this isn't a correct proof. But here's how I guessed it. $2^{48}\mod97=2^{-48}\mod97=M$ $\therefore M^2=1$ Therefore I guessed that M=1.

(Technically, M can also be 96, but, ...)

(More formal method) $\begin{array}{l} &2^7\mod97&=&128\mod97&=&31\\ =&2^{14}\mod97&=&31^2\mod97&=&961\mod97&=&-9\\ =&2^{28}\mod97&=&(-9)^2\mod97&=&81\mod97&=&-16\\ =&2^{42}\mod97&=&(-9)(-16)\mod97&=&144\mod97&=&47\\ =&2^{43}\mod97&=&94\mod97&=&-3\\ =&2^{48}\mod97&=&-3(32)\mod97&=&-96\mod97&=&1 \end{array}$

This is not the most formal method, because of the use of negative numbers.