Calculators are useless

The last digit in the base is $\color{#3D99F6}{3}$ , Now a base with 3 in its unit digit and having some magnitude of exponent, can have $3, 9, 7, 1$ only in its unit place. So, the expression will have either:

• A $1$ in its unit digit if the exponential is of the type $\color{#D61F06}{4n}$ .

• A $3$ in its unit digit if the exponential is of the type $4n+1$ .

• A $9$ in its unit digit if the exponential is of the type $4n+2$ .

• A $7$ in its unit digit if the exponential is of the type $4n+3$ .

Now any factorial above $3!$ is divisible by $4$ and thus will be of the type $\color{#D61F06}{4n}$ , and so the unit digit in the given expression will be a $\boxed{\color{#3D99F6}{1}}$ .

We observe that $987654321!$ is a multiple of $100$ so it is a multiple of $4$ . Now we see that the last digit of the base is $3$ whose powers' last digit repeat in the cycle $3,9,7,1$ since $987654321! \equiv 0 \left(\text{mod 4}\right)$ the units digit of the expression is $\color{#3D99F6}{\boxed{1}}$