Find all of the prime factors of $ 20! $ Could someone please give me a hint on how to go about this? I have pretty much missed one week of summer school so I don't know if there is a more elegant way to arrive at a solution other that using the fundamental theorem of arithmetic. 
 A: Hint: Can a prime bigger than $20$ be a divisor of $20!$? If there was a "prime" bigger than $20$ that divided $20!$, what can you say about this "prime"? Is it really "prime"?
A: $$
20! = 1\times2 \times3 \times4\times5 \times6\times 7\times8 \times9 \times10 \times11 \times12 \times 13\times 14\times 15\times 16\times 17\times 18\times 19 \times 20.
$$
The prime factors of this number are the prime factors of these $20$ factors.  For example $15$ is one of those, and its prime factors are $3$ and $5$.  So we get all prime numbers less than or equal to these numbers.  Those are $2,3,5,7,11,13,17,19$.
A: \begin{align}
20! & = 2 \times 3 \times 4 \times 5 \times 6 \times \cdots \times 19 \times 20 \\
    & = (2) \times (3) \times (2^2) \times (5) \times (2 \times 3) \times \cdots
        \times (19) \times (2^2 \times 5) \\
    & = 2^{18} \times 3^8 \times 5^4 \times 7^2 \times 11 \times 13 \times
        17 \cdot 19
\end{align}
A: Hint.  Start with the obvious factorisation of $20!$, then proceed to the prime factorisation.
A: Use the Legendre theorem:
Let 
$$
n!=p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k} \cdots.
$$
Then 
$$
\alpha_k=\sum_{i>0}\left[ \frac{n}{p_k^i}\right].
$$
In your case $20!=2^{\alpha_2} 3^{\alpha_3} \cdots 19^{\alpha_{19}}$.
Then you may calculate every $\alpha_k$ separately, for example
$$
\alpha_2=\left[ \frac{20}{2}\right]+\left[ \frac{20}{4}\right]+\left[ \frac{20}{8}\right]+\left[ \frac{20}{16}\right]+\left[ \frac{20}{32}\right]=10+5+2+1+0=18.
$$
