How do you simplify ${{i}^{25}}$ ?

Hint: For answering this question we will use the basic concept of complex number which is given as $i=\sqrt{-1}$ and the basic formulae ${{i}^{4q+r}}={{i}^{r}}$. Now by applying these concepts we will come to a conclusion for this question.

Complete step by step answer:
Now considering from the question we have to find the value of ${{i}^{25}}$ .
From the basic concept of complex numbers we know that $i=\sqrt{-1}$ and from these have one basic formulae ${{i}^{4q+r}}={{i}^{r}}$ this can be derived by substituting values in place of $q$ and $r$ . Let us do that. By substituting $r=0$ and $q=1$ we will have ${{i}^{4}}={{\left( \sqrt{-1} \right)}^{4}}$ which can be further simplified as ${{i}^{4}}={{\left( -1 \right)}^{2}}=1$ which is equal to ${{i}^{0}}=1$ .
Now we are going to use this formula in our question we can write $25=4\left( 6 \right)+1$ and we can also say that ${{i}^{25}}={{i}^{4\left( 6 \right)+1}}$ by comparing it with the formulae we observe that $q=6$ and $r=1$ .
So we will have ${{i}^{25}}={{i}^{4\left( 6 \right)+1}}=i$ .

Hence we can conclude that the value of ${{i}^{25}}$ is equal to $i$ that is $\sqrt{-1}$ .

Note: We should be sure with our calculations and concepts while answering questions of this type. In complex numbers we have many other concepts like these which we can prove similarly as shown here.