= Solution
<Mathematical induction> consists of an <induction base case> $P(1)$ and an <inductive step> $P(n)\Rightarrow P(n+1)$ for every <positive integer> $n$. To derive it from the <well-ordering principle>, suppose the <set> of counterexamples were nonempty and take its least element $n$. The base case excludes $n=1$. If $n>1$, minimality gives $P(n-1)$, and the step then gives $P(n)$, a <contradiction>. Hence there are no counterexamples.
For the stated <integer congruence>, the base case is $9\equiv2\pmod7$. If $9^n\equiv2^n\pmod7$, multiplication and $9\equiv2\pmod7$ give
$$
9^{n+1}\equiv9\cdot2^n\equiv2^{n+1}\pmod7,
$$
completing the induction.
The purported shifted-sum proof has a correct conditional successor calculation, but \b[the base case is missing and false]: at $n=1$ it would assert $1=1+126$. An <inductive step> without a true <induction base case> cannot establish any instance.
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