In the principle of Mathematical Induction, why do we take the base case as $P(1)$ only? I kinda understand the logic and motivation behind the proof, but what bothers me is the fact, why is the base case (the first statement that we write) is always $P(1)$ when we are proving a proposition is true for all $\mathbb{N} $? Why can't it be anything other than $1$? What about $P(5)$ or something? Does it change anything?
 A: A typical induction hypothesis (not always) is saying that if $k$-th statement is true, then $(k+1)$-th statement is true. Imagine that you have some dominoes, the $(k+1)$-th dominoes fall because $k$-th domino hits it. Now, if you started proven that the first is fallen, now you can use the induction hypothesis to conclude that the $2$nd domino falls as well and so on. Then your conclusion is valid for $n \ge 1$.
If you only started pushing from the $5$-th dominos, the first $4$ dominos could still be standing. Your conclusion can only be made for $n \ge 5$.
Also, try proving $n! \ge 2^n, \forall n \ge 4$ by induction, it is likely that you will use a different base case. 
A: The short answer is that we don't always start with $P(1).$ It depends on what we're trying to prove. Sometimes we'll have multiple base cases, or start with a greater/lesser base case.
The idea of mathematical induction is more that we build up on the base case(s) with the induction step(s), thereby handling infinitely-many cases without having to tackle them one by one.
Added: For an example where we start with a larger base case, see here. For an example where we want multiple base cases, see here. Here is an example of a proof with multiple induction steps. Here's an outline of a proof that induces in both directions from the base case $P(0),$ proving a claim for all integers.
A: We need to construct proof for a statement $P(n)$ for all $n\ge 1$
As basis case, we need to show $P(1)$ is true.
Then in induction  step, we prove $P(k) \implies P(k+1)$.
Hence  $P(1) \implies P(2), P(2) \implies P(3) \cdots \cdots$
If you take $P(5)$ as basis case, then 
$P(5) \implies P(6), P(6) \implies P(7) \cdots \cdots$
Thus in case you take $P(5)$ as basis case, You need to prove your statement explicitly for $n=1,2,3,4$
