This is not necessarily true, and in fact you can get a counterexample where each $\lambda_i=0$. Consider $V=c_0$ and let $N_i=\sum_{j=0}^i j$ be the $i$th triangular number. Define $T:c_0\to c_0$ by $$T(x)_k=\begin{cases} 0 & \text{if }k=N_i\text{ for some }i \\ 2x_{k-1} & \text{otherwise}\end{cases}$$

That is, $T$ is twice the right shift, except that the $N_i$th coordinates of $T(x)$ are declared to be $0$ for each $i$. Clearly $T$ is bounded, with $\|T\|=2$.

If $V_i$ is the space of sequences supported on $[N_i,N_{i+1})$, then each $V_i$ is $T$-stable and $T$ is nilpotent on $V_i$. As $\bigoplus V_i$ is dense in $c_0$, we conclude that $T$ satisfies your hypotheses with $\lambda_i=0$ for all $i$. Now let $x\in c_0$ be the sequence such that $x_{N_i}=1/i$ for each $i$ and all other coordinates are $0$. Then for any $n$, $T^n(x)$ will have $(N_i+n)$th coordinate $2^n/i$ whenever $i\geq n$. In particular, $\|T^n(x)\|\to\infty$.

how bigthe powers can get $\endgroup$ – Yemon Choi Jun 15 '15 at 0:34