I would like someone to take a look for the following argument-maybe somebody find it useful.
1. Let us recall the so called *bar resolution* for an algebra $A$. It is the sequence of the form
$$...P_n \stackrel{b'}{\longrightarrow} P_{n-1} \stackrel{b'}{\longrightarrow} ... \stackrel{b'}{\longrightarrow}P_0 \stackrel{m}{\longrightarrow} A$$
where $m$ is the multiplication map and where $P_n=A \otimes A^{op} \otimes A^{\otimes n}$ which is isomorphic to $A^{\otimes (n+2)}$ via the map $$a \otimes a' \otimes a_1 \otimes ... \otimes a_n \mapsto a \otimes a_1 \otimes ... \otimes a_n \otimes a'.$$
There is a one-to-one correspondence between $A-A$ bimodule structures and left $A \otimes A^{op}$ module structures and we view each $P_k$ as left $A \otimes A^{op}$ module.
One can check that even if $A$ is noncommutative, operator $b'$ is always $A \otimes A^{op}$ linear (unlike operator $b$).
Since each $P_n$ is of the form $A \otimes A^{op} \otimes (something)$ we conclude that each $P_n$ is free $A \otimes A^{op}$ module, therefore there are projective in particular.
Exactness of this sequence is provided by the existence of an operator $s$ which acts as follows
$$s(a_1 \otimes ... \otimes a_n):=1 \otimes a_1 \otimes ... \otimes a_n.$$
It is straighforward that
$$b's+s'b=1$$ so the above complex (in the category of left $A \otimes A^{op}$ modules) is contractible so it is exact. This is how we get *bar resolution* (beeing a projective resolution for $A$). The general theory tells us that Hochschild homology $H_*(A,M)$ may be computed by tensoring this resolution with $M$.

2. Denote by $D_n$ the subspace of $A^{\otimes n}$ generated by the expressions $a_1\otimes a_2 \otimes ... \otimes a_n$ where some $a_i=1$ and let $\overline{A^{\otimes n}}$ denote the quotient $A^{\otimes n}/D_n$. We construct projective resolution $(\overline{P_n})_n$ for $A$ where $\overline{P_n}:=A \otimes A^{op} \otimes \overline{A^{\otimes n}}$. The fact that this is projective resolution is proved similarly as for bar resolution provided we can show that $b'$ and $s$ ,,pass to the quotient''---this can be directly and easily verified.

3. General homological algebra tells us that the projective resolution is unique up to chain homotopy: there is some subtlety---slightly more is true. If we take two projective resolutions $(P_n)_n$ and $(\overline{P_n})_n$ there is up to chain homotopy *one and only one* chain map $P_* \to \overline{P_*}$ and $\overline{P_*} \to P_*$. But there is natural chain map $P_* \to \overline{P_*}$ namely the projection map $p$. General theory provides us with the chain map going opposite way: $q:\overline{P_*} \to P_*$ and from uniqueness we have that $pq$ and $qp$ are chain homotopic to identities.

4. From this we get that the maps $1_M \otimes p: M \otimes_{A \otimes A^{op}} P_* \to M \otimes_{A \otimes A^{op}} \overline{P_*}$ and $1_M \otimes q: M \otimes_{A \otimes A^{op}} \overline{P_*} \to M \otimes_{A \otimes A^{op}} P_*$ when composed are chain homotopic to identities, therefore both of them are quasi-isomorphisms.

5. Let us recall the argument that the (standard) bar resolution computes Hochschild homology. We consider the map $$\alpha: M \otimes_{A \otimes A^{op}} A \otimes A^{op} \otimes A^{\otimes n} \to M \otimes A^{\otimes n},$$ $$ \alpha(x \otimes a \otimes a' \otimes a_1 \otimes ... \otimes a_n):=a'xa \otimes a_1 \otimes ... \otimes a_n $$
One shows that such $\alpha$ is an isomorphism (in particular it is the chain map: it intertwines $b'$ from the bar resolution with the standard Hochschild boundary $b$).

6. We can repeat this argument for $\overline{A^{\otimes n}}$ in place of $A^{\otimes n}$:

a) $\alpha$ indeed passes to quotients---denote this map by $\overline{\alpha}$;

b) $b$ is well defined as a mapping $M \otimes \overline{A^{\otimes n}} \to M \otimes \overline{A^{\otimes (n-1)}}$.

The above statements may be verified easily. So we proved that our *normalized* projective resolution computes the homology of the *normalized* complex. But being a projective resolution for $A$ it must also compute ordinary Hochschild homology. Therefore we get that the homology of the normalized complex is isomorphic to Hochschild homology.

7. We want slightly more: we want our projection $M \otimes A^{\otimes n} \to M \otimes \overline{A^{\otimes}}$ to induce the isomorphism in homology. But this map is the composition $\overline{\alpha} \circ (1_M \otimes p) \circ \alpha^{-1}$---the composition of isomorphism, quasi isomorphism and isomorphism, therefore it is quasi isomorphism.

8. For *co* homology the argument is similar modulo the following:

a) normalized cochains (i.e. maps defined on $\overline{A^{\otimes n}}$) may be viewed as ordinary cochains, i.e. maps on $A^{\otimes n}$

b) applying $Hom(-,M)$ functor to the quotient map gives the inclusion.

So the inclusion of normalized complex into the whole Hochschild complex induces isomorphism in cohomology.

9. We would like to have analogous result for cyclic theory. Since cyclic homology (or cohomology) cannot be computed using bar resolution, we cannot repeat everything for cyclic theory. However we can start defining cyclic theory using normalized complexes and develop the theory in the context of normalized complexes, until we arrive at another, *normalized* version of Connes $IBS$ sequence. Then we have a commutative diagram where the first row is normalized IBS sequence, the second is ordinary IBS sequence and vertical maps comes from the inclusion. Using the inductive argument (our rows starts with some zeros, which allows us to start the induction) and five lemma we obtain that the inclusion of cyclic cochains into all cochains induces the isomorphism in cohomology.

cyclic cohomology-- but I may have misremembered this last comment $\endgroup$ – Yemon Choi Jul 23 '16 at 13:11