Proof by Induction: I lost points on this proof because its incomplete and not descriptive enough. any suggestions?
If $a_i$ $\geq$ $0$, i $\geq$ 1 
then, (1+$a_1$)(1+$a_2$).....(1+$a_n$)$\geq$ 1 + $a_1$ + $a_2$ + ..... + $a_n$
Proof by Induction: 
case(n=1), we have (1+$a_1$) $\geq$ 1 + $a_1$ is true.
suppose true for n $\geq$ 1.
Then (1+$a_1$)(1+$a_2$)....(1+$a_n$)(1+$a_{n+1}$) $\geq$ (1+$a_1$ + $a_2$ + ...$a_n$)(1+$a_{n+1}$)
By the inductive hypothesis,
1+$a_1$+$a_2$+.....+$a_n$+$a_{n+1}$+$a_{n+1}$($a_1$+....+$a_n$) $\geq$ 1 + $a_1$ +....+$a_{n+1}$
Therefore by induction,
(1+$a_1$)(1+$a_2$).....(1+$a_n$)$\geq$ 1 + $a_1$ + $a_2$ + ..... + $a_n$ true $\forall$n$\in$$\mathbb{Z^+}$
 A: I think it is fine but not very reader-friendly. Especially in the context of an exam, think of making the corrector's life easier, especially:


*

*don't make calculation jumps, it is not immediately clear how lines above and below "By the inductive hypothesis" in your question relate. Prefer chaining (in-)equalities, such as : $A \ge B = B' = B'' \ge \text{(by assumption X)  } C$, it is usually easier to follow 

*mark clearly the "anchors" of the theories/tools/theorems you are using. In the context of induction, that would be:


*

*Suppose property $P_n$ is true for $n$ : $A_n \ge B_n$

*(calculations)

*which yields $A_{n+1} \ge B_{n+1}$, i.e. $P_{n+1}$ is true

*since $P_0$ is true and $P_n \implies P_{n+1}$, $P_n$ is true for every $n \ge 0$



Of course this may seem very verbose, and the amount of detail to be provided depends on the level of the class. But keep in mind that the purpose of an exam/exercise is to show to the instructor that you know your course/how to crack a problem so make it clear.
PS: the same subject is also discussed in this question: Solution verification: Prove by induction that $a_1 = \sqrt{2} , a_{n+1} = \sqrt{2 + a_n} $ is increasing and bounded by $2$. Maybe someone should take the time to write a community wiki Q/A some time.
Late edit: I have tried to provide some basic help on induction proof writing here : Proof writing: how to write a clear induction proof?
A: You could do
$$\color{red}{(1+a_1)\cdot\ldots\cdot(1+a_n)}(1+a_{n+1})\stackrel{\text{Ind. Hyp.}}\ge\color{red}{(1+a_1+\ldots+a_n)}(1+a_{n+1})$$
It is thus enough to show
$$(1+a_1+\ldots+a_n)(1+a_{n+1})\ge1+a_1+\ldots a_n+a_{n+1}\iff$$
$$\color{green}{1+a_1+\ldots a_n+a_{n+1}}+a_1a_{n+1}+a_2a_{n+1}+\ldots a_na_{n+1}\ge\color{green}{1+a_1+\ldots+a_{n+1}}\iff$$
$$a_1a_{n+1}+\ldots+ a_na_{n+1}\ge 0$$
and since the last inequality is trivial we're done.
A: Assuming that the statement is true for arbitrary $n$: Letting $n \mapsto n+1$ in the Statement one obtains:
$(1+a_1)...(1+a_n)(1+a_{n+1}) = (1+a_1)...(1+a_n) + (1+a_1)...(1+a_n)a_{n+1} \geq 1+a_1+...+a_{n+1}$.
Now it holds $a_{n+1} \leq (1+a_1)...(1+a_n)a_{n+1} $ because $a_i \geq 0$; hence:
$(1+a_1)...(1+a_n) + a_{n+1} \geq 1+a_1+...+a_{n+1}$
Therefore this Statement remains true for every natural number $n$.
