Can we prove that $26^n$ always ends in the same two digits?

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Prove that, for all positive integral values of \(m\), \(x-1\) is a factor of \(x^m - 1\).

Prove that, if \(n\) is an integer greater than \(1\), then \(26^n - 26\) is an odd multiple of \(50\).

Hence, or otherwise, show that, if \(26^n\) is evaluated for any integral value of \(n\) greater than \(1\), then the last two digits of the result are independent of \(n\).

[The phrase ‘integral values’ means the same as ‘integer values’. ]