 Solution

Euler's theorem states that $a^{\varphi (m)}\equiv 1(\mathrm{mod}m)$ whenever $\gcd (a,m)=1$, where $\varphi (m)$ counts the invertible residue classes. For each prime-power divisor, the same theorem gives $a^{\varphi (p_i^{e_i})}\equiv 1(\mathrm{mod}{p}_i^{e_i})$. Since the exponent defined in the source is a multiple of every one of these exponents, raising that congruence to the corresponding integer power givesThe prime powers are pairwise coprime, so their product divides $a^{\lambda (m)}-1$, proving the congruence modulo $m$.

For the displayed composite integer,Thus every integer coprime to 561 satisfies $\boxed{a^{560}\equiv 1(\mathrm{mod}561)}$. This is a Carmichael number. The source's prime-power least common multiple need not be the minimal universal exponent for all moduli; for powers of two its definition can exceed the usual Carmichael function. No minimality claim is needed here.