Let $T$ be a complete first order theory.
Assuming $\lambda \geq |L(T)| + \aleph_0$, is it obvious that $I(T, \lambda) \leq 2^{\lambda}$ ? The question was spurred by looking at Vaught's conjecture on wikipedia. I don't have much model theory knowledge so it might be obvious and I'm just being dumb or it might be using technology that I haven't covered yet. Or is this perhaps only true when $\lambda = \aleph_0$ ?
The only thing I can think of is to count the number of non isomorphic models based on which types they satisfy, but I don't think that quite works.
In the setting that you're describing there are at most $2^\lambda$ $L$-structures, and hence certainly no more than $2^\lambda$ isomorphism classes of models of $T$.
Fix a $\lambda \geq |L|+\aleph_0$, and let $M$ be a set of cardinality $\lambda$.
For each $n$-ary function symbol $f$ we must interpret $f$ as a function $f : M^n \to M$. There are exactly $\lambda^{(\lambda^n)} = 2^\lambda$ many such functions to choose from. Similarly, for each $n$-ary relation symbol we have $2^\lambda$ options to choose from, and for each constant symbol we have only $\lambda$ options to choose from.
Thus for each symbol in the language we have at most $2^\lambda$ choices. Since there are at most $\lambda$ symbols in the language, the total number of choices we have to make is at most $(2^\lambda)^\lambda = 2^{\lambda \times \lambda} = 2^\lambda$.