It becomes clear once you understand what $p-1$ and ECM are doing. I'll start with $p-1$, since it is simpler.
Suppose the number you want to factor, $n$, is the product of two primes: $p$ and $q$. If we knew $p-1$, we could get $p$ from $n$ by doing
$$
\gcd(a^{p-1} - 1 \bmod n, n) = p.
$$
This works because by working modulo $n$, we're working "in parallel" modulo $p$ and modulo $q$. And modulo $p$, we get $a^{p-1} - 1 = 0 \bmod p$ (for nonzero $a$) by Fermat's little theorem. So $a^{p-1} - 1 \bmod n$ is going to be 0 modulo $p$, but not modulo $q$, and the $\gcd$ gives us a factorization of $n$.
However, we don't really know $p-1$ — if we did, we wouldn't be looking for $p$ — and that puts a dent in our plans. There is an alternative route, however: if we can find a multiple of $p-1$, that works too:
$$
a^{k(p-1)} - 1 \bmod p \equiv (a^{p-1})^k - 1 \bmod p \equiv 1^k - 1 \bmod p = 0 \bmod p.
$$
So what Pollard's $p-1$ does is select an exponent that is likely to contain every single factor of $p-1$. An easy way to do this, that works as long as $p-1$ only has small-ish factors, is simply to make the exponent the product of all primes: $2\cdot3\cdot5\cdot7\ldots\cdot B_1$, where $B_1$ is our upper bound. This, however, misses power-of-prime factors, such as $2^2$ or $3^5$. So another exponent that catches everything is the factorial of $B_1$: that one is certain to catch every possible factor up to $B_1$!
Some observations about $p-1$ leads us to ECM. $p-1$ is the order of the group formed by the integers modulo $p$, denoted by $\mathbb{F}_p$. Further, if the order of $\mathbb{F}_p$ only has small factors, we can find $p$ by the aforementioned exponentiation. It turns out this generalizes to other mathematical groups. In particular, the group of points in an elliptic curve modulo $p$ is parameterized by two additional numbers $a$ and $b$:
$$
E_{a,b} = \{ (x, y) : y^2 = x^3 + a x + b \pmod{p} \}
$$
Each curve (parameterized by a and b) modulo $p$ has an order ($\#E$, analogous to $p-1$ above) in the range
$$
p + 1 - 2\sqrt{p} < \#E < p + 1 + 2\sqrt{p}.
$$
Further, for each order in that interval, there is an $(a,b)$ tuple that yields it.
Suppose we pick a random elliptic curve $E_{a,b}$, and run it modulo $n$ (again, "in parallel" in both $p$ and $q$). Picking a random point in the curve, $P$, and multiplying it by $\#E$ results in the point at infinity (the elliptic curve analogous of 0):
$$
\#E\cdot P = \infty
$$
This special point, usually represented as $(0,1)$, triggers divisions by $0 \bmod p$ but not $\bmod~q$, and that nets us a factor. The logic for picking the exponent is the same as $p-1$: every prime and prime power up to $B_1$, and hope for a hit.
Besides the factorial, another good way to pick the exponent is the lowest common multiple of every integer up to $B_1$; this yields every prime power lower than $B_1$, and is much smaller than the factorial of $B_1$.
