For any \[n\in N,{{i}^{8n+1}}\] is?

Hint: In this problem, we have to find, for any \[n\in N\], the value of \[{{i}^{8n+1}}\]. We are given that n belongs to natural numbers, we know that natural numbers start from 1. We can first split the given expression using the exponent rule. We can then simplify them using imaginary rules such as \[{{i}^{2}}=-1,{{i}^{4}}=1,{{i}^{3}}=-i\]. After using the exponent rule we will get the power as multiples of four, whose value will be 1. We can then simplify and find the answer.

Complete step by step answer:
Here we have to find the value of \[{{i}^{8n+1}}\] for any \[n\in N\].
We can now write the given imaginary expression using the exponent rule \[{{a}^{x+y}}={{a}^{x}}\times {{a}^{y}}\],
\[\Rightarrow {{i}^{8n}}\times {{i}^{1}}\] ……. (1)
We should know that there are some values and rules of imaginary terms, they are
\[{{i}^{2}}=-1,{{i}^{4}}=1,{{i}^{3}}=-i\] …… (2)
 We can now write the first term in (1) by writing it in multiples of 4, we get
\[\Rightarrow {{i}^{4\times 2n}}\times i\]
We know that any number in multiples of four will give the same value as \[{{i}^{4}}=1\] and hence n can be any natural number which will give only the multiples of 4, so we can write it as,
\[\Rightarrow 1\times i=i\]
Therefore, the value of \[{{i}^{8n+1}}\] for any \[n\in N\] is \[i\].

Note: We should always remember the imaginary numbers rules such as \[{{i}^{2}}=-1,{{i}^{4}}=1,{{i}^{3}}=-i\]. We should also remember the exponent rules which can be used in these types of problems. If the power in the imaginary term is in multiples of 4 then we will get the value as 1 as the answer.