Look at the base, not in the exponent

Suppose that $a,b,c,d$ are the exponent. So, the expression can be expressed as $2^a + 3^b +4^c + 5^d$ .

Notice that $2^a$ and $4^c$ are even, so the sum of this is also an even number.

And, note that $3^b$ and $5^d$ are odd, so the sum of this is even number.

Thus, $2^a + 3^b +4^c + 5^d$ is even.

Therefore, the smallest prime factor is 2.

Haha nice trick question! And the title is appropriately given!

We can also do it by Binomial Theorem,

We can write the expression as,

$2^{3^{4^{5}}} + (2+1)^{4^{5^{2}}} + 4^{5^{2^{3}}} + (4+1)^{2^{3^{4}}}$

Now grouping the factors of $2$ we get,

$2^{3^{4^{5}}} + (2)^{4^{5^{2}}}... + 4^{5^{2^{3}}} + (4)^{2^{3^{4}}}... + {1}^{4^{5^{2}}} + {1}^{2^{3^{4}}}$

or,

$2^{3^{4^{5}}} + (2)^{4^{5^{2}}}... + 4^{5^{2^{3}}} + (4)^{2^{3^{4}}}... + 2$

Hence the smallest factor is 2.