>I have to prove by MI that, x^5 - 1 is a multiple of 5.

Sorry, but it fails at x = 2.
Is there a typo?

>My analysis is as follows:
>
>1.If some no say x has to be a muliple of 5, then x/5 = n
>where n=1,2,3 etc.

No, (x^5 - 1) is a multiple of 5, not x.

>2. In the prob above mentioned, while following the first
>step of MI,and substituting x with 1,we get the answer as 0.
>Now, 0 cannot be effectively called a "multiple" of 5.

But 0 is a multiple of 5.

Consider: "Is 18 a multiple of 6?"
Yes, because 18/6 = 3. That is, 18 is divisible by 6 with remainder 0.
Try that with "Is 0 a multiple of 5?"

Alternate approach: "Is 18 a multiple of 6?"
Yes, because 18 = 6k, for some integer k (namely, 3).
Is 0 = 5k for some integer k?