Your formula for $\mathrm P(M=k)$ is wrong because the random variables $U_i$ for $i\ne k$ are not independent.
In fact, $[M=k]=[U_k\gt V_k]$ with $V_k=\max\limits_{i\ne k}U_i$ hence $\mathrm P(M=k)=\mathrm P(V_k\lt U_k)$. Since the random variables $U_k$ and $V_k$ are independent, integrating this identity with respect to the distribution of $U_k$ yields $ \mathrm P(M=k)=\frac1{T_k}\int_0^{T_k}\mathrm P(V_k\lt u)\mathrm du=\frac1{T_k}\sum\limits_{i=1}^k\int_{T_{i-1}}^{T_i}\mathrm P(V_k\lt u)\mathrm du, $ with the notation $T_0=0$. If $T_{i-1}\lt u\lt T_i$ with $i\leqslant k$, then $U_j\lt u$ with full probability for every $j\leqslant i-1$ hence $[V_k\lt u]=[\forall j\geqslant i,j\ne k,U_j\lt u]$ and $ \mathrm P(M=k)=\frac1{T_k}\sum\limits_{i=1}^k\int_{T_{i-1}}^{T_i}\prod\limits_{i\leqslant j\leqslant n,j\ne i}\frac{t}{T_j}\mathrm dt, $ that is, $ \mathrm P(M=k)=\sum\limits_{i=1}^k\frac{T_i^{n-i+1}-T_{i-1}^{n-i+1}}{(n-i+1)T_iT_{i+1}\cdots T_n}=\sum\limits_{i=1}^k\frac{q_i-q_{i-1}}{n-i+1}, $ where, for every $0\leqslant i\leqslant n$, $ q_i=\frac{T_i^{n-i}}{T_{i+1}T_{i+2}\cdots T_n}. $ One sees that $\mathrm P(M=k)\leqslant\mathrm P(M=k+1)$ and that the inequality is strict as soon as $T_i\lt T_{i+1}$ since $ q_{i+1}=q_i\left(\frac{T_{i+1}}{T_i}\right)^{n-i}. $
A simple construction of
$M$ which might help to explain why the result is true is as follows. Consider a random index
$1\leqslant I\leqslant n$ whose distribution is characterized by the fact that, for every
$1\leqslant i\leqslant n$,
$ \mathrm P(I\leqslant i)=q_i,\qquad \mathrm P(I=i)=q_i-q_{i-1}. $ Then, conditionally on
$I$,
$M$ is uniformly distributed in the set
$\{I,I+1,\ldots,n\}$.
This yields directly the explanation you are after. Choose $1\leqslant k\leqslant n$. Then:
- If $i\leqslant k$, then $\mathrm P(M=k\mid I=i)=\mathrm P(M=k+1\mid I=i)$.
- For $i=k+1$, $\mathrm P(M=k\mid I=k+1)=0$ while $\mathrm P(M=k+1\mid I=k+1)\gt0$.
- If $i\gt k+1$, then $\mathrm P(M=k\mid I=i)=0=\mathrm P(M=k+1\mid I=i)$.
Summing these contributions yields $\mathrm P(M=k)\lt\mathrm P(M=k+1)$. This also yields once again the difference $ \mathrm P(M=k+1)-\mathrm P(M=k)=\mathrm P(M=I=k+1)=\frac{\mathrm P(I=k+1)}{n-k}=\frac{T_{k+1}^{n-k}-T_k^{n-k}}{(n-k)T_{k+1}\cdots T_n}. $