prove no identity element in the given Cayley Table of three elements Prove that the operation in the following Cayley table has no identity element:
$$
\begin{array}{c|ccc}
\hline
* & u & v & w \\
\hline
u & u & w & w \\
v & v & v & v \\
w & w & u & v \\
\hline
\end{array}
$$
The only way I can think is just by checking it one by one. But I don't think that be considered a proof.
 A: Since there are only three elements, you can certainly "just" check each case. Indeed, that's exactly what you should probably do: "Proof by cases" is a perfectly legitimate method of proof: 
Consider each element as a separate "case" (there are only three to consider). For each element $x \in \{u, v, w\}$, find a counterexample which shows that $x$ cannot be the identity. 
For example, suppose we test the case where $x = w$: From the table, we have that $w*w = v.\;$ This means $w$ cannot be an identity, since if $x = w$ were the identity, we must have $w \times w = w$. $\;\checkmark$
Do the same for each of $u$ and $v$, and you're done.

One observation, as noted below in the comments: See if you can prove that if an identity element $x$ exists, we would need to have one column for element $x$ replicate the left-most column, and the corresponding row for $x$ replicate the top most row, the "header" row. 
Example where there exists an identity element $u$: 
$$
\begin{array}{l}
\text{Example with identity u} \\
\begin{array}{c|ccc}
\hline
* & u & v & w \\
\hline
u & u & v & w \\
v & v & w & u \\
w & w & u & v \\
\hline
\end{array}
\end{array}
$$
A: I will in a way just reword the second part of the other answer by amWhy.
You notice the identity element from a Cayley table, if there is one, because in this case the line and the column of the table labeled by that element both show the identity function on the underlying set (not just a bijection).
In your example
$$
\begin{array}{c|ccc}
\hline
* & u & v & w \\
\hline
u & \color{red}{u} & w & w\\
v & \color{red}{v} & v & v\\
w & \color{red}{w} & u & v\\
\hline
\end{array}
$$
no line shows $u,v,w$ in this order while the column in red, labeled by $u$, does. Indeed here $u$ is a right-identity only and not a left identity.
Pay attention because if an identity element exists, than it is unique; instead left and right identity elements may exist and not be unique. Further:


*

*if one identity element exists, than no other element can be a left nor right identity;

*if two or more left (right) identities exist, than no element can be a right (left) identity.

