Winning All Levels in a Game There are $L$ levels in a game. In each turn of the game, you go through each level one by one and try to complete it. The goal is to complete all levels of the game. The probability of completing any one of the $L$ levels in  a single turn is $p$. If you complete a particular level at a previous turn then that progress is saved and you don't have to complete it in any successive turns. Even if you fail to complete any level at a particular turn, then the turn continues with the other levels(you don't go to a new turn). So in each turn, you try all the $L$ levels. On average, how many turns do you have to play the game to complete all the levels?
This was my approach. Let $N_k$ be the average number of turns you need to play the game in order to complete any $k$ of those $L$ levels($0\le$ $k$ $\le$ $L$). I write the following recurrence relation(from which I can easily calculate $N_L$, the desired answer, since $N_0=0$).
$(N_k+1)[1-(1-p)^{L-k}]+(N_{k+1}+1)(1-p)^{L-k}=N_{k+1}$
This is because if you win any one of the remaining $L-k$ levels in the next turn, you have taken $N_k+1$ turns to complete $k+1$ levels and if you lose all, you need $N_{k+1}+1$ turns to complete $k+1$ levels. Is this recurrence correct?
Is there any loophole in my logic?
 A: You recurrence relation may be correct, but here is a different perspective. 
The number of tries required to complete one level, is independent of all the other levels, and is distributed in some manner you can figure out (probability $p$ to finish in one try, $(1-p)p$ in a second try, you can see the distribution coming out of this).
Introduce $L$ new random variables, say $R_1, ... , R_L$,where each $R_i$ is an independent identically distributed variable, which stores the number of tries required to finish the $i$th level.
Now the total number of tries you will need is $R_1 + ... + R_L$, and the expectation value $E[R_1 + ... + R_L] = \sum E[R_i]$, can you figure out $E[R_i]$ now with the distribution you have got for $R_i$? This method is much easier than evaluating a recurrence relation, especially of the kind you have put out.
EDIT: Your question seems to have changed, then I will refer you to the answer below, which is the same a the one I am getting.
A: Consider this formulation of your problem: When you play a level, you have two options: win the level with probability $p$ (and ascend to the next level) or die with probability $1-p$ (and spend another turn). Consider the following question: How many times do you die on average before you win $L$ times?
Clearly this number is $L$ times the number of times you die on average before you win one level, since beating each level takes the same effort on average. If you win with probability $p$, then you'll expect to win the first time after $1/p$ tries (the expected value of the geometric distribution), which means you'll have died $1/p-1$ times on average. So to beat the entire game, you'll have to repeat this process $L$ times, the number of deaths will be $L/p-L$ in total, and the number of turns you need is therefore $L/p-L+1$.
A: Let $q = 1-p$ 
We can view the process as a "group knockout" tournament of $L$ players,
where only $1$ out of each group of $q$ makes it to the next round until we get the "winner"
Working backwards, in the finals, we must have $q$ players, in the semi-finals, $q^2$ players and so on.
Thus if $n$ rounds are needed, $q^n = L,\;\; or\;\; n = \frac {L}{log\;\; q}$
But here we need to "knock out" the winner also (complete the last level), which needs an additional $\frac{1}{p}$ "ghost matches" !
Thus  answer $ = \frac {L}{log\;\; q} + \frac{1}{p}$ 
