Suppose $A$ is nilpotent, and $B = c_0I_n + c_1A + \cdots + c_{m-1}A^{m-1}$, show $\det(B) = 0$ iff $c_0 = 0$ Let $A$ be an $n\times n$ matrix, with real entries. If $A$ is nilpotent, so $A^m = 0$ for some $m \ge 1$, $B = c_0I_n + c_1A + \cdots + c_{m-1}A^{m-1}$, where $c_i \in R$ for each $i$, show that $\det(B) = 0$ if and only if $c_0 = 0$.

My attempt: 

So, $c_0 = 0$ implies $\text{det}(B) = 0$ is obvious: if $c_0 = 0$, then $\det(B) = \det((c_1 + c_2A + \cdots + c_{m-1}A^{m-2})A) = \det(c_1 + c_2A + \cdots + c_{m-1}A^{m-2})\cdot \det(A) = 0,$ since $\det(A) = 0$ if $A$ is nilpotent.

I'm not sure how to tackle the reverse direction. If $\text{det}(B) = 0$, we know: $B$ is not invertible, we know that $\det(c_0I_n + c_1A + \cdots + c_{m-1}A^{m-1}) = 0$. But I don't know where to go from here! A hint in the right direction would be appreciated.
If $N$ is nilpotent of order $k$, then $I-N$ is invertible because
$$
    (I-N)(I+N+N^{2}+\cdots+N^{k-1})=I-N^{k}=I.
$$
Now suppose $A$ is nilpotent and further suppose $c_0 \ne 0$. Then
$$
         c_0 I + c_1 A + c_2 A^{2}+\cdots + c_{m-1}A^{m-1} = c_0(I-N),
$$
where $N$ is nilpotent. Hence, the above is invertible if $c_0 \ne 0$. However, the above is definitely not invertible if $c_0=0$ because, in that case, the above is nilpotent.
Do the following:

Consider $B^n$ the n-th power of B, with $n\geq m$ 

$B^n = (c_0 I + c_1 A + \cdots + c_{m-1} A^{m-1}) \cdot \ldots \cdot (c_0 I + c_1 A + \cdots + c_{m-1} A^{m-1})$

$B^n = c_0^n I^n + \sum_{2n \geq i\geq n} \alpha_i A^i  = c_0^n I$

With $\sum_{2n \geq i\geq n} \alpha_i A^i$ I mean that $B^n$ is equal to $c_0^n I$ plus some powers of $A$ multiplied by an scalar but the exponent of those powers of $A$ is grater than $m$ therefore $A^i=0$ and $B^n = c_0^n I$.

Now use what you know $0 = det(B)^n = det(B^n) = det(c_0^n I) = c_0^k det(I) = c_0^k$ therefore $c_0 = 0$.