To evaluate $5^2,$ we write it out with multiplication signs and then compute the product:

\begin{align*} 5^2 = \underbrace{5 \times 5 }_{\large \text{$2$ times}} = 25 \end{align*}

The square of $9$ is the same as $9^2.$

To evaluate $9^2,$ we write it out with multiplication signs and then compute the product:

\begin{align*} 9^2 = \underbrace{9 \times 9 }_{\large \text{$2$ times}} = \bbox[3pt,Gainsboro]{\color{blue}81} \end{align*}

We need to evaluate $7^2.$

To evaluate $7^2,$ we write it out with multiplication signs and then compute the product:

\begin{align*} 7^2 = \underbrace{7 \times 7 }_{\large \text{$2$ times}} = 49 \end{align*}

We need to evaluate $3^3.$

To evaluate $3^3,$ we start by writing it out with multiplication signs: \[ 3^3 = \underbrace{3 \times 3 \times 3}_{\large \text{$3$ times}} \]

Now, we compute the product. Since $3 \times 3 = 9,$ we have \begin{align*} \underbrace{3 \times 3}_{\large\color{blue} 9} \times 3 &= {\color{blue}9} \times 3. \end{align*}

Finally, since $9\times 3=27,$ we have $3^3=27.$

We need to evaluate $7^3.$

To evaluate $7^3,$ we start by writing it out with multiplication signs: \[ 7^3 = \underbrace{7 \times 7 \times 7}_{\large \text{$3$ times}} \]

Now, we compute the product. Since $7 \times 7 = 49,$ we have \[ \underbrace{7 \times 7}_{\large\color{blue} 49} \times 7 = {\color{blue}49} \times 7. \]

Finally, we work out ${\color{blue}49} \times 7.$

\begin{align*} \begin{array}{ccccc} & & \!\!\!\! \underset{\color{blue}6}{} \!\!\!\! & \\ & & \!\!\!\! 4 \!\!\!\! & \!\!\!\! 9 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & & & \!\!\!\! 7 \!\!\!\! \\ \hline & \!\!\!\! 3 \!\!\!\! & \!\!\!\! 4 \!\!\!\! & \!\!\!\! 3 \!\!\!\! \end{array}\\ & \end{align*}

Since $49\times 7=343,$ we have $7^3=\bbox[3pt,Gainsboro]{\color{blue}{343}}.$

Here, ${\color{blue}3}^{\color{red}1}$ is in exponent form. The base is $\color{blue}3$ and the exponent is ${\color{red}1}.$

This means that $\color{blue}3$ is taken as a factor only once. Therefore, \[ 3^1 = \bbox[3pt,Gainsboro]{\color{blue}3}. \]

Here, ${\color{blue}23}^{\color{red}1}$ is in exponent form. The base is $\color{blue}23$ and the exponent is ${\color{red}1}.$

This means that $\color{blue}23$ is taken as a factor only once. Therefore, \[ 23^1 = 23. \]

Here, ${\color{blue}0}^{\color{red}4}$ is in exponent form. The base is $\color{blue}0$ and the exponent is ${\color{red}4}.$

This means that ${\color{blue}0}$ is taken as a factor ${\color{red}4}$ times. Therefore, \[ 0^4 = \underbrace{0 \times 0 \times 0 \times 0}_{\large \text{$4$ times}} = \bbox[3pt,Gainsboro]{\color{blue}0}. \]

Here, ${\color{blue}1}^{\color{red}4}$ is in exponent form. The base is $\color{blue}1$ and the exponent is ${\color{red}4}.$

This means that ${\color{blue}1}$ is taken as a factor ${\color{red}4}$ times. Therefore, \[ 1^4 = \underbrace{1 \times 1 \times 1 \times 1}_{\large \text{$4$ times}} = 1. \]