Solve T(n) = T(n-1)*n with repeated substitution
This is the rucursive formula for $n$ factorial.

Define the formula $T(n)$ for $n \geq 0$ as $T(n) = nT(n-1)$. By repeated substitutions observe that 
$$ T(0) = 1$$
$$ T(1) = 1\times T(0)=1 \times 1$$
$$ T(2) = 2 \times T(1)=2\times 1\times 1$$
$$ T(3) = 3\times T(2)=3\times 2\times 1\times 1$$
$$\dots$$
$$T(n) = nT(n-1)=n\times (n-1)(n-2)\dots 1\times 1 $$
Thus by simple observation you can easily guess the pattern of product of the first $n$ positive integers.

For a complete proof you need to prove that $T(n) = n!$. This is easily done by induction. For $n=0$, $T(n)=1 = 0!$. Assume that formula is true for $n-1$. Then $T(n)=nT(n-1)=n(n-1)!=n!$ which completes the proof. So $T(n)$ is $\Theta(n!)$.

There is no a standard method to solve such kind of recurrence relations, but you usually guess the explicit form of the formula and proof it, for example, using induction or other technique or method.

As an exercise try to guess and prove explicit formulas for 


$T(0) = 1, T(n)=2T(n-1)$
$T(0) = 1, T(n) = nT(n-2)$, for only even $n$s.