<s>[INST] By knowing, that $a_{n+1}=(n+3)a_n$, how can I find $a_n$? $$a_1=1$$
$$a_{n+1}=(n+3)a_n$$
How can I get to the answer of this, which is:
$$a_n=\frac{(n+2)!}{6}$$
 [/INST] write it as follows 
$$\frac { { a }_{ n+1 } }{ { a }_{ n } } =n+3\\ \frac { { a }_{ 2 } }{ { a }_{ 1 } } =4,\frac { { a }_{ 3 } }{ { { a } }_{ 2 } } =5,\frac { { a }_{ 4 } }{ { a }_{ 3 } } =6,...\frac { { a }_{ n } }{ { a }_{ n-1 } } =n+2\\ \\ \frac { { a }_{ 2 } }{ { a }_{ 1 } } \cdot \frac { { a }_{ 3 } }{ { { a } }_{ 2 } } \cdot \frac { { a }_{ 4 } }{ { a }_{ 3 } } \cdot ...\cdot \frac { { a }_{ n } }{ { a }_{ n-1 } } =\frac { \left( n+2 \right) ! }{ 6 } $$
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