Five. (Full marks 15 points) Expand new numbers according to the following rules:
Given two numbers $a$ and $b$, a new number can be expanded according to the rule $c = ab + a + b$. Any two numbers among $a$, $b$, and $c$ can then be used to expand another new number according to the rule, and so on. Each expansion of a new number is called an operation.
Starting with the numbers 1 and 4,
(1) Find the maximum new number that can be obtained by performing the above rule three times;
(2) Can the new number 1999 be expanded according to the above rule? Explain your reasoning.
(1) The first time, you can only get $1 \times 4+4+1=9$.
Since the goal is to get the maximum new number, the $\cdots$ time, take 4 and 9, to get $4 \times 9+4+9=49$.
Similarly, the second time, take 9 and 49, to get
$9 \times 49+9+49=441$.
(2) $\because a=a \cdot a+b=(a+1)(b+1)-1$,
$$
\therefore a+1=(a+1)(b+1).
$$

Taking numbers $a, c$ to get the new number
$$
\begin{array}{l}
d=(a+1)(a+1)-1=(a+1)(b+1)(a+1)-1 \\
=(a+1)^{2}(b+1)-1, \\
\therefore d+1=(a+1)^{2}(b+1).
\end{array}
$$

Taking numbers $b, a$ to get the new number
$$
\begin{aligned}
e & =(b+1)(c+1)-1=(b+1)(a+1)(b+1)-1 \\
& =(b+1)^{2}(a+1)-1.
\end{aligned}
$$

Let the final new number be $x$. Then it can always be expressed as
$$
x+1=(a+1)^{m} \cdot(b+1)^{n}.
$$

Where $m, n$ are integers.
$$
\begin{array}{l}
\text { When } a=1, b=4, x+1=2^{n'} \times 5^{n}. \\
x \because 1999+1=2000=2^{4} \times 5^{3}.
\end{array}
$$

Therefore, 1999 can be obtained through the aforementioned rules.