Idea is correct, proof lacks rigor, coefficient of $t$ in $\det(I+tA)$ The goal of this exercise is to find the coefficient of $t$ in the polynomial $\det(I+tA)$ where $I, A \in Mat_n(\mathbb R)$.
I've thought about this long and hard as I am combinatorically challenged (it's the most difficult branch of mathematics for certain) and I'm pretty sure I have a convincing argument, but it feels handwavey and lacks rigor. I'd like someone to review it and possibly suggest way to make it less verbal and more rigorous.
My argument uses Laplace expansion, initially using the first column of $I+tA$.
$\det(I+tA) = (1+ta_{1,1})\begin{vmatrix}1+ta_{2,2} & ta_{2,3} & \dots & ta_{2,n}\\ta_{3,2} & 1+ta_{3, 3} & \dots & ta_{3,n}\\ \vdots & \vdots & \dots & \vdots \\ ta_{n, 2} & ta_{n, 3} & \dots & 1+ta_{n,n}\end{vmatrix} -ta_{2,1}\det(B_2) + ta_{3,1}\det(B_3)-\dots+(-1)^{n+1}ta_{n, 1}\det(B_n)$
Where $B_i$ are some sub matrices with accordance to Laplace.
Notice that since we used the first column in our original expansion, the first row of every $B_i$ is just $\begin{pmatrix}ta_{1,2} \dots ta_{1,n}\end{pmatrix} = t\begin{pmatrix}a_{1,2} \dots a_{1,n}\end{pmatrix} $
if we define $B_i'$ to be exactly the same as $B_i$ except the first row is divided by $t$, then by determinant rules $\det(B_i) = t\det(B_i')$, and so:
$-ta_{2,1}\det(B_2) + ta_{3,1}\det(B_3)-\dots+(-1)^{n+1}ta_{n, 1}\det(B_n) = -t^2a_{2,1}\det(B_2') + t^2a_{3,1}\det(B_3')-\dots+(-1)^{n+1}t^2a_{n, 1}\det(B_n')$
and they only contain non linear terms of $t$. So they don't matter to us at all. We can just focus on the first determinant.
But now we can just use the exact same argument on $\begin{vmatrix}1+ta_{2,2} & ta_{2,3} & \dots & ta_{2,n}\\ta_{3,2} & 1+ta_{3, 3} & \dots & ta_{3,n}\\ \vdots & \vdots & \dots & \vdots \\ ta_{n, 2} & ta_{n, 3} & \dots & 1+ta_{n,n}\end{vmatrix}$! The only thing that will matter is the first element and its leading minor. The others can be discarded as they are higher order of $t$. 
We will use it again and again until at the end we are only left with $\prod_{i=1}^{n}(1+ta_{i,i})$. The answer is hidden somewhere in this product.
Since we are only interested in linear terms, we are not allowed to multiply $ta_{i,i}$ with $ta_{j,j}$. So every $ta_{i,i}$ can only be multiplied by $1$. So overall we should have $ta_{1,1} + ta_{2,2} + \dots ta_{n,n} = tTrace(A)$.
Is this result correct? is this "proof" valid? It feels too wishy washy and verbal.
 A: The result is correct. I think your argument is fine, although it is not very amenable to be written nicely. 
Here is a shorter proof. 
If $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$ counting multiplicities, then the eigenvalues of $I+tA$ are $1+t\lambda_1,\ldots,1+t\lambda_n$. Thus
$$
\det(I+tA)=(1+t\lambda_1)\cdots(1+t\lambda_n)=t^n\lambda_1\cdots\lambda_n+\cdots+t(\lambda_1+\cdots+\lambda_n)+1.
$$
So the coefficient of $t$ is 
$$
\lambda_1+\cdots+\lambda_n=\operatorname{Tr}(A). 
$$
We also see from above that the coefficient of $t^n$ is $\det A$. 
A: Here is a different argument. Let the columns of $A$ be $C_i$. Then the determinant viewed as a function of the sequence of columns is multilinear.  Let $E_i$ be the columns of the identity matrix. Then you have 
$$\det(E_1+tC_1, \cdots , E_n+tC_n)$$ now by mulitlinearily this can be expanded like an binomial product, and thus the coefficient of $t$ is 
$$\sum \det(E_1, \cdots , C_i,\cdots , E_n)$$  and it is easy to see that 
$$ \det(E_1, \cdots , C_i,\cdots , E_n)=a_{ii}.$$  
