Invariant subspaces

What are the invariant subspaces of $f$ an endomorphism on $V$ over $\mathbb{C}$ with basis $e_1, \ldots, e_n$ such that $f(e_1)=e_1, f(e_i)=e_i+e_{i-1}$ for $i\gt 1$?

I am thinking the invariant subspaces must just be $\langle e_1, \ldots, e_k\rangle$ for $1\le k\le n$. But I don't know how to prove or disprove that there are no other invariant subspaces.

-
Suppose an invariant subspace contains $a_1e_1+a_2e_2+\ldots+a_ke_k$, with $a_k \neq 0$, and show that it then contains all of $\langle e_1, \ldots, e_k\rangle$. – Chris Eagle Oct 20 '12 at 12:44
I see. Keep applying $f-id$ to $a_1e_1+\cdots+a_ke_k$ and eventually get $a_k e_1$, and hence $e_1$ since $a_k\neq 0$. Applying $f-id$ one time less and get $a_{k-1}e_1+a_ke_2$ so that way I have $e_2$. Repeat to generate all of $e_1, \cdots, e_k$. – Chuwei Zhang Oct 20 '12 at 13:07
By the way, if I come up with an answer to my own question after getting a hint, like this, should I answer my own question or just leave the answer in the comment fields? – Chuwei Zhang Oct 20 '12 at 13:09
You should write and accept your own answer. – Chris Eagle Oct 20 '12 at 13:17

Keep applying $f−id$ to $a_1e_1+⋯+a_ke_k$ and eventually get $a_ke_1$, and hence $e_1$ since $a_k\neq0$. Applying $f−id$ one time less and get $a_{k−1}e1+a_ke_2$ so that way I have $e_2$. Repeat to generate all of $e_1,⋯,e_k$.