even vs odd

Removing the trailing zeroes of any whole number $k$ is the same as dividing it by the largest power of ten that is its factor; the size of this power of ten is, in turn, determined by the number of 2's or the number of 5's, whichever is smaller, in the prime factorization of $k$ .

For $p$ a prime, the prime factorization of $n!$ will contain at least one $p$ for every multiple of $p$ in its expansion, so for $n \ge 2$ , $n! \color{#333333} = 1 \cdot \color{#20A900}2 \color{#333333} \cdot 3 \cdot \color{#20A900}{4} \color{#333333} \cdot \color{#3D99F6}5 \color{#333333} \cdot \color{#20A900}{6} \color{#333333} \cdot 7 \cdot \ldots \cdot n$ will clearly always contain more 2's than 5's. Therefore the statement in the problem is true for any $n \ge 2$ .

To be more specific, for $p$ a prime, every multiple of $p$ in the expansion of $n!$ will contribute one $p$ to the prime factorization of $n!$ ; however every multiple of $p^2$ will contribute two $p$ 's rather than one, and in general every multiple of $p^k$ will contribute $k$ $p$ 's. Thus the number of $p$ 's in the prime factorization of $n!$ is given by

$\displaystyle\sum_{i=1} \left\lfloor \dfrac{n}{p^i} \right\rfloor$

Since the largest power of ten that is a factor of $n!$ is limited by the number of 5's in its prime factorization, removing the trailing zeroes of $n!$ will remove all its 5's and an equal number of 2's. Thus the remaining number will not only be even, but divisible by $2^m$ , where

$m=\displaystyle\sum_{i=1} \left\lfloor \dfrac{n}{2^i} \right\rfloor\ - \displaystyle\sum_{i=1} \left\lfloor \dfrac{n}{5^i} \right\rfloor$

For example, if $n=37$

$\begin{aligned} m&=\displaystyle\sum_{i=1} \left\lfloor \dfrac{n}{2^i} \right\rfloor\ - \displaystyle\sum_{i=1} \left\lfloor \dfrac{n}{5^i} \right\rfloor \\ \\ &= \left( \left\lfloor \dfrac{37}{2} \right\rfloor\ + \left\lfloor \dfrac{37}{4} \right\rfloor\ + \left\lfloor \dfrac{37}{8} \right\rfloor\ + \left\lfloor \dfrac{37}{16} \right\rfloor\ + \left\lfloor \dfrac{37}{32} \right\rfloor\ \right) - \left( \left\lfloor \dfrac{37}{5} \right\rfloor\ + \left\lfloor \dfrac{37}{25} \right\rfloor\ \right) \\ \\ &= (18 + 9 + 4 + 2 + 1) - (7 + 1) \\ \\ &= 26 \end{aligned}$

so after removing the (eight) trailing zeroes of $37!$ , the remaining number will still be divisible by $2^{26}$ .

We can notice that for $n\geq 5$ $n! = a_0a_1a_2\cdots\times a_n \times 10^n \\ n! = a_0a_1a_2\cdots a_n \times \left({\color{#3D99F6}2^n\times 5^n}\right)$ it's says that the number of trailing zeroes depend on the prime factors of 2 and 5 in case base is 10. Since the number 2's is greater than 5's in $n!$ and even after removing the trailing $n!$ still has some 2's say $2^r$ . Then we can conclude that $n! = 2\left(p_0p_1p_3\cdots p_n\right)\cdot 2^{r-1} = 2k$ Hence, the last digits non-zeros digit of $n!$ is always even.