The last digit is my target

Note that for $n!$ when $n>4$ has the prime factors of $2$ and $5$ and product of 2 and 5 are responsible for the trailling zeros of $n!$ . More in general for $n\geq 5$

$\begin{aligned} & n! = 1\times 2\times 3\times \cdots 5\times \cdots \times n\\& n! = a \times{\color{#3D99F6} (2^{k}\times 5^{k})} =a\times{\color{#3D99F6}(10)^{k}}\quad (a,k) \in\mathbb Z^{+}\end{aligned}$ Note : The exponents of 2 and 5 must be checked and equalised to that of exponent of 5 since the exponents of 2 is greater than that of 5 for $n!$

For $n > 5$

$n! = 1 \times 2 \times 3 \times 4 \times 5 \times \ldots \times n = 1\color{#3D99F6}0 \color{#333333} \times 3 \times 4 \times \ldots \times n = 2\color{#3D99F6}0 \color{#333333} \times 2 \times 3 \times \ldots \times n$

Any integer multiplied by $\overline{m\color{#3D99F6}0}$ will end in $\color{#3D99F6}0$ . Where $m$ is a positive integer.

Note: $5! = 12\color{#3D99F6}0$