= Solution
Assume the incubation times are <independent> observations from the same upper-<truncated distribution>. Conditional on lying below $M$, the <gamma distribution> density is $g(t;a,s)/G(M;a,s)$ on its support. With $t_{(n)}=\max_i t_i$, the <likelihood function> is
$$
\boxed{L(a,s,M)=\frac{\prod_{i=1}^n t_i^{a-1}\exp(-t_i/s)}{s^{na}\Gamma(a)^nG(M;a,s)^n}
\mathbf1\{0<t_i<M\text{ for every }i\}.}
$$
In particular, the normalizing factor $G(M;a,s)^{-n}$ cannot be omitted. On the admissible region the <log-likelihood> is
$$
\ell=(a-1)\sum_i\log t_i-\frac{\sum_i t_i}{s}
-na\log s-n\log\Gamma(a)-n\log G(M;a,s).
$$
The <profile likelihood> eliminates the <nuisance parameters> $a,s$ by fitting them separately at each candidate endpoint:
$$
L_p(M)=\sup_{a>0,s>0}L(a,s,M).
$$
For fixed $a,s$ and $M>t_{(n)}$,
$$
\frac{\partial\ell}{\partial M}=-n\frac{g(M;a,s)}{G(M;a,s)}<0.
$$
Equivalently, for $t_{(n)}<M_1<M_2$, every nuisance-parameter fit has $L(a,s,M_1)>L(a,s,M_2)$. Taking suprema shows that $L_p$ is nonincreasing. If the nuisance maximum at $M_2$ is attained and finite, evaluating its maximizing parameters at $M_1$ shows strict decrease there as well. Values below the largest observation are inadmissible. Thus <upper-endpoint estimation in a truncated distribution> gives the usual boundary estimate
$$
\boxed{\widehat M=t_{(n)}.}
$$
There is a support-convention qualification: the literal strict support $t<M$ excludes $M=t_{(n)}$, so it gives a supremum approached as $M\downarrow t_{(n)}$, rather than an attained <maximum-likelihood estimator>. Using the equivalent closed-endpoint density convention for a continuous distribution yields the usual boundary estimate above. The argument presumes a meaningful finite nuisance fit; degenerate samples can instead have an unbounded density <likelihood>. Since $M$ changes the support, regular interior-parameter asymptotic formulas are not automatically applicable.
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