Is this language NP Hard? $L=\{$$($$m$,$w$,$n$$)$| $m$ is an encoding of a non-deterministic Turing machine, $w$ is any word/string in the closure of alphabet, i.e. $w\in\Sigma^*$, $n$ is any positive integer, i.e. $n\in\Bbb{Z}$ and $n&gt;0$, and $m$ accepts $w$ for a maximum of $n$ moves/steps/operations $\}$

Alphabet is $\Sigma=\{0,1\}$

If this language is NP Hard then what reduction proves it?

My effort to answer this question myself is:

If $L$ is NP Hard then every NP language reduces to $L$ in polynomial time deterministically.

More formally $L\in\Bbb{NPH}\implies\forall L'\in\Bbb{NP}:\;L'\le_PL$

Let $L'$ be any arbitrary NP language.

According to the assumption that $L'\in\Bbb{NP}$ by definition of the complexity class $\Bbb{NP}$ exists non-deterministic Turing machine $m$ that decides $L'$ in polynomial time.

This means that for every word/string $w\in\Sigma^*$:

If $w\in L'$ then $m$ accepts $w$ for a maximum of $c\cdot N^k$ moves/steps/operations

AND

If $w\notin L'$ then $m$ rejects $w$ for a maximum of $c\cdot N^k$ moves/steps/operations

Where $c$ and $k$ are constant non-negative integers and $N$ is the length of $w$.

If $L'\le_PL$ then exists deterministic Turing machine $m'$ that computes function $f:\Sigma^*\rightarrow\Sigma^*$ in polynomial time so that for every word/string $w\in\Sigma^*$:

$w\in L'\iff f(w)\in L$

Let $m'$ be deterministic Turing machine that in the beginning $m'$ inserts the encoding of $m$ before input word/string $w$, which takes constant time $\mathcal{O}(1)$, because the encoding of $m$ is constant word/string and thus the length of the encoding of $m$ is constant non-negative integer, then $m'$ computes the length of the input word/string $w$ by counting all symbols of $w$, which suppose to take linear time $\mathcal{O}(N)$ where $N$ is the length of the input word/string $w$, then compute the power of the computed length by the constant non-negative integer $k$, which takes polynomial time, then multiply the result by the constant non-negative integer $c$, which also takes polynomial time, and at last append the result after $mw$ and the content of the tape of $m'$ will be $(m,w,c\cdot N^k)$ where $n=c\cdot N^k$

After doing all these actions $m'$ halts.

$m'$ is indeed deterministic Turing machine and runs in polynomial time and $w\in L'\iff (m,w,c\cdot N^k)\in L$ followed by the definitions of complexity class $\Bbb{NP}$ and non-deterministic Turing machines.

Hence by definition $L'\le_PL$ because of $m'$

Conclusion: For arbitrary language $L'$ I showed that $L'\le_PL$ holds.

Therefore every NP language reduces to $L$ in polynomial time deterministically and by definition $L$ is NP Hard.

What I wrote above suppose to be a proof that $L$ is NP Hard, but the proof doesn't reduce already proven NP Hard language to $L$, so I think that I did some mistake.
Indeed, this is the prototypical NP-complete language.

Recall what NP-hardness means: For all languages $L'$ in NP, there is a polynomial time reduction from $L'$ to $L$. 
Normally, when one wants to prove NP-hardness, one takes a language $L''$ which is itself already known to be NP-hard and establishes a reduction from $L''$ to $L$.

But not so in this case! It is much more natural to directly construct a reduction from $L'$ to our $L$, only using the fact that $L'$ belongs to NP.