#### Answer

$n²$ is odd because:
$n$ is odd and can be write as $2k + 1$
$n² = 2(2k² + 2k) + 1$

#### Work Step by Step

If $n$ is an odd integer. $n$ can be write as:
$n = 2 k + 1$
(with $k$ being one integer number)
$k \in \mathbb{Z}$
so
$\begin{split}
n^2 & = (2k + 1)^2 \\
& = (2k)^2 + 2.2k.1 + 1^2 \\
& = 4k^2 + 4k + 1 \\
& = 2(2k^2 + 2k) + 1 \\
\end{split}$
Since $2k^2 + 2k$ is an integer, then $2(2k^2 + 2k) + 1$ is also odd, by the definition of odd numbers.